blob_id string | repo_name string | path string | length_bytes int64 | score float64 | int_score int64 | text string | is_english bool |
|---|---|---|---|---|---|---|---|
3c41155cbab16bb91e4387aa35d49e1832ce5b72 | codedsun/LearnPythonTheHardWay | /ex32.py | 525 | 4.25 | 4 | #Exercise 32: Loops and List
the_count = [1,2,3,4,5]
fruits = ['apples','oranges','pears','apricots']
change = [1,'pennies',2,'dimes',3,'quaters']
#for loop in the list
for number in the_count:
print("This is %d count"%number)
for fruit in fruits:
print("A fruit of type %s"%fruit)
for i in change:
print("I got %r"%i)
#Building empty list and then appending it
element=[]
for i in range(0,6):
print("Adding the %d in the list"%i)
element.append(i)
for i in element:
print("Element %d"%i)
| true |
8c3788b2fc95ce08c9dc6165b6f5de91e7d7d448 | semihyilmazz/Rock-Paper-Scissors | /Rock Paper Scissors.py | 2,933 | 4.40625 | 4 | import random
print "********************************************************"
print "Welcome to Rock Paper Scissors..."
print "The rules are simple:"
print "The computer and user choose from Rock, Paper, or Scissors."
print "Rock crushes scissors."
print "Paper covers rock."
print "and"
print "Scissors cut paper."
print "You will play the computer...the first to reach 3 wins wins the match."
print "Good luck!"
print "************************************************************"""
user_win = 0
computer_win= 0
user_name= raw_input("Whats Your Name? ")
while True:
computer_choice = ""
user_choice = raw_input("Please choose (R) Rock, (P) Paper, or (S) Scissors: ").upper()
random_number = random.randint(1,3)
if random_number == 1:
computer_choice = "R"
if random_number == 2:
computer_choice = "P"
if random_number == 3:
computer_choice = "S"
if computer_choice == "R":
print("""
_______
---' ____)
(_____)
(_____)
(____)
---.__(___)
The computer chose rock.""")
elif computer_choice == "P":
print("""
_______
---' ____)____
______)
_______)
_______)
---.__________)
The computer chose paper.""")
elif computer_choice == "S":
print("""
_______
---' ____)____
______)
__________)
(____)
---.__(___)
The computer chose scissors.""")
print ""
if (computer_choice == "R" and user_choice == "P") or \
(computer_choice == "P" and user_choice == "S") or \
computer_choice == "S" and user_choice == "R":
print "You win!"
user_win +=1
print user_name, user_win
print "computer=", computer_win
elif (computer_choice == "R" and user_choice == "S") or \
(computer_choice == "P" and user_choice == "R") or \
(computer_choice == "S" and user_choice == "P"):
print "You lose!"
computer_win +=1
print user_name, user_win
print "computer=", computer_win
else:
print "Ties"
print user_name, user_win
print "computer=", computer_win
if computer_win == 3:
print user_name,"LOST"
user_win =0
computer_win = 0
delay = raw_input("You want to play another game Y/N ? ")
if delay == "y":
continue
else:
break
elif user_win == 3:
print user_name,"WON"
user_win = 0
computer_win = 0
delay = raw_input("You want to play another game Y/N ? ")
if delay == "y":
continue
else:
break
| true |
9a6fe2ffbb280d239d50001e637783520b8d4aef | songseonghun/kagglestruggle | /python3/201-250/206-keep-vowels.py | 272 | 4.1875 | 4 | # take a strign. remove all character
# that are not vowels. keep spaces
# for clarity
import re
s = 'Hello, World! This is a string.'
# sub = substitute
s2 = re.sub(
'[^aeiouAEIOU ]+', #vowels and
# a space
'', #replace with nothing
s
)
print(s2)
| true |
b5363f3d6c5b6e55b3d7607d055a266b215c28ab | dianamenesesg/HackerRank | /Interview_Preparation_Kit/String_manipulation.py | 2,979 | 4.28125 | 4 | """
Strings: Making Anagrams
Alice is taking a cryptography class and finding anagrams to be very useful. We consider two strings to be anagrams of each other if the first string's letters can be rearranged to form the second string. In other words, both strings must contain the same exact letters in the same exact frequency For example, bacdc and dcbac are anagrams, but bacdc and dcbad are not.
Alice decides on an encryption scheme involving two large strings where encryption is dependent on the minimum number of character deletions required to make the two strings anagrams. Can you help her find this number?
Given two strings, and , that may or may not be of the same length, determine the minimum number of character deletions required to make and anagrams. Any characters can be deleted from either of the strings.
For example, if and , we can delete from string and from string so that both remaining strings are and which are anagrams.
Function Description
Complete the makeAnagram function in the editor below. It must return an integer representing the minimum total characters that must be deleted to make the strings anagrams.
makeAnagram has the following parameter(s):
a: a string
b: a string
Input Format
The first line contains a single string, .
The second line contains a single string, .
Constraints
The strings and consist of lowercase English alphabetic letters ascii[a-z].
Output Format
Print a single integer denoting the number of characters you must delete to make the two strings anagrams of each other.
Sample Input
cde
abc
Sample Output
4"""
import math
import os
import random
import re
import sys
# Complete the makeAnagram function below.
def makeAnagram(a, b):
for character_a in a:
if character_a in b:
a = a.replace(character_a,"",1)
b = b.replace(character_a,"",1)
num_deleted = len(a) + len(b)
return num_deleted
#print(makeAnagram("jxwtrhvujlmrpdoqbisbwhmgpmeoke", "fcrxzwscanmligyxyvym"))
"""
Alternating Characters
A Shashank le gustan las cadenas donde los caracteres consecutivos son diferentes. Por ejemplo, le gusta , mientras que no le gusta. Dada una cadena que solamente contiene caracteres y , él quiere cambiarla a una cadena que le guste. Para hacerlo, solo se le permite borrar los caracteres en la cadena.
Tu tarea es encontrar la mínima cantidad requerida de borrados.
Formato de Entrada La primera linea contiene un enter que quiere decir el número de casos de prueba. Luego siguen lineas , con una cadena en cada linea.
Formato de Salida Imprimie la mínima cantidad requerida de pasos en cada caso de prueba.
Restricciones
Ejemplo de Entrada
5
AAAA
BBBBB
ABABABAB
BABABA
AAABBB
Ejemplo de Salida 00
3
4
0
0
4
"""
def alternatingCharacters(s):
#return len([1 for x in range(len(s)-1) if s[x]==s[x+1]])
case = 0
for i in range(len(s)-1):
if s[i] == s[i+1]:
case += 1
return case
#alternatingCharacters('AAABBBAABB')
| true |
c5d1e2535e6e8d9ac0218f06a44e437b3358049b | michaelschung/bc-ds-and-a | /Unit1-Python/hw1-lists-loops/picky-words-SOLUTION.py | 543 | 4.3125 | 4 | '''
Create a list of words of varying lengths. Then, count the number of words that are 3 letters long *and* begin with the letter 'c'. The count should be printed at the end of your code.
• Example: given the list ['bat', 'car', 'cat', 'door', 'house', 'map', 'cyst', 'pear', 'can', 'spike'], there are exactly 3 qualifying words.
'''
words = ['bat', 'car', 'cat', 'door', 'house', 'map', 'cyst', 'pear', 'can', 'spike', 'ace', 'cpe']
count = 0
for word in words:
if len(word) == 3 and word[0] == 'c':
count += 1
print(count)
| true |
c9a7bdeb85a94afc6b17c2b6ddf3be0d78de6206 | yangju2011/udacity-coursework | /Programming-Foundations-with-Python/turtle/squares_drawing.py | 755 | 4.21875 | 4 | import turtle
def draw_square(some_turtle):
i = 0
while i<4: # repeat for squares
some_turtle.forward(100)
some_turtle.right(90)
i=i+1
def draw_shape():#n stands for the number of edge
windows = turtle.Screen()#create the background screen for the drawing
windows.bgcolor("red")
brad = turtle.Turtle() #move around, def __init__ stand for initialize
brad.shape("turtle")#https://docs.python.org/2/library/turtle.html#turtle.shape
brad.color("white","blue")#two color, first outline, second filled; can't do three colors
brad.speed(5)
for i in range(0,36):
brad.right(10)
draw_square(brad)
windows.exitonclick()
draw_shape()
| true |
1bb3d699d6d1e7ef8ea39a5cdc7d55f79ac4ed3d | theelk801/pydev-psets | /pset_lists/list_manipulation/p5.py | 608 | 4.25 | 4 | """
Merge Lists with Duplicates
"""
# Use the two lists below to solve this problem. Print out the result from each section as you go along.
list1, list2 = [2, 8, 6], [10, 4, 12]
# Double the items in list1 and assign them to list3.
list3 = None
# Combine the two given lists and assign them to list4.
list4 = None
# Replace the first 3 items in list 3 with the numbers 4, 3, 12.
# Create a variable list5. Merge list3 and list4 to create a list containing no duplicates.
list5 = None
# Take a look at your printed statements to see the evolution of your lists with each step of this problem.
| true |
d7aa917a00b2f582c7b15c8b40de0462ca274a66 | theelk801/pydev-psets | /pset_classes/fromagerie/p6.py | 1,200 | 4.125 | 4 | """
Fromagerie VI - Record Sales
"""
# Add an instance method called "record_sale" to the Cheese class. Hint: You will need to add instance attributes for profits_to_date and sales (i.e. number of items sold) to the __init__() method in your Cheese class definition BEFORE writing the instance method.
# The record_sale method should do the following:
### Add the sale's profits to profits_to_date attribute.
### Add the number sold to the running total in the sales attribute.
### Subtract the number sold from the stock attribute.
### Check the stock (Hint: Call the check_stock method within the record_sale method.)
# Once finished, call record_sale on each sale from the sales_today dict below and print out the name of the cheese, whether it has a low stock alert, the remaining stock value, the running total of units of that cheese sold, and the running total of your profits to date. It should look something like this:
"""
Parmigiano reggiano - Low stock alert! Order more parmigiano reggiano now.
Remaining Stock: 14
Total Sales: 10
Profits to Date: 20
"""
sales_today = {provolone: 8,
gorgonzola: 7, mozzarella: 25, ricotta: 5, mascarpone: 13, parmigiano_reggiano: 10, pecorino_romano: 8} | true |
1859b100bd9711dfa564f8d3a697202bc42d96c2 | theelk801/pydev-psets | /pset_conditionals/random_nums/p2.py | 314 | 4.28125 | 4 | """
Generate Phone Number w/Area Code
"""
# import python randomint package
import random
# generate a random phone number of the form:
# 1-718-786-2825
# This should be a string
# Valid Area Codes are: 646, 718, 212
# if phone number doesn't have [646, 718, 212]
# as area code, pick one of the above at random
| true |
278cfd9000b17eaf249eebad07b11d078e934c35 | theelk801/pydev-psets | /pset_classes/bank_accounts/solutions/p2.py | 1,377 | 4.21875 | 4 | """
Bank Accounts II - Deposit Money
"""
# Write and call a method "deposit()" that allows you to deposit money into the instance of Account. It should update the current account balance, record the transaction in an attribute called "transactions", and prints out a deposit confirmation message.
# Note 1: You can format the transaction records like this - "Deposit: <starting balance>, <ending balance>"
# Note 2: You can format the deposit confirmation message like this - "$<amount> deposit complete."
class Account():
def __init__(self, name, account_number, initial_amount):
self.name = name
self.acc_num = account_number
self.balance = initial_amount
self.transactions = {}
int_rate = 0.5
def acc_details(self):
print(f'Account No.: {self.acc_num}\nAccount Holder: {self.name}\nInterest Rate: {self.int_rate}\nBalance: ${self.balance}\n')
def deposit(self, amount):
start_amount = self.balance
self.balance += amount
end_amount = self.balance
label = 'Deposit'
self.transactions.update({label: [start_amount, end_amount]})
print(f'${amount} deposit complete.')
alejandra = Account('Alejandra Ochoa', 554951, 20000)
alejandra.deposit(500)
print(f'${alejandra.balance}') # 20500
print(alejandra.transactions) # {'Deposit': [20000, 20500]}
alejandra.acc_details() | true |
9d9d044b0bff764cc3a8f4e31634f1138f05e2b4 | Arin-py07/Python-Programms | /Problem-11.py | 888 | 4.1875 | 4 | Python Program to Check if a Number is Positive, Negative or 0
Source Code: Using if...elif...else
num = float(input("Enter a number: "))
if num > 0:
print("Positive number")
elif num == 0:
print("Zero")
else:
print("Negative number")
Here, we have used the if...elif...else statement. We can do the same thing using nested if statements as follows.
Source Code: Using Nested if
num = float(input("Enter a number: "))
if num >= 0:
if num == 0:
print("Zero")
else:
print("Positive number")
else:
print("Negative number")
The output of both programs will be the same.
Output 1
Enter a number: 2
Positive number
Output 2
Enter a number: 0
Zero
A number is positive if it is greater than zero. We check this in the expression of if. If it is False, the number will either be zero or negative. This is also tested in subsequent expression.
| true |
5b428c1de936255d8a787f6939528031fa3ddc9b | Arin-py07/Python-Programms | /Problem-03.py | 1,464 | 4.5 | 4 | Python Program to Find the Square Root
Example: For positive numbers
# Python Program to calculate the square root
# Note: change this value for a different result
num = 8
# To take the input from the user
#num = float(input('Enter a number: '))
num_sqrt = num ** 0.5
print('The square root of %0.3f is %0.3f'%(num ,num_sqrt))
Output
The square root of 8.000 is 2.828
In this program, we store the number in num and find the square root using the ** exponent operator. This program works for all positive real numbers. But for negative or complex numbers, it can be done as follows.
Source code: For real or complex numbers
# Find square root of real or complex numbers
# Importing the complex math module
import cmath
num = 1+2j
# To take input from the user
#num = eval(input('Enter a number: '))
num_sqrt = cmath.sqrt(num)
print('The square root of {0} is {1:0.3f}+{2:0.3f}j'.format(num ,num_sqrt.real,num_sqrt.imag))
Output
The square root of (1+2j) is 1.272+0.786j
In this program, we use the sqrt() function in the cmath (complex math) module.
Note: If we want to take complex number as input directly, like 3+4j, we have to use the eval() function instead of float().
The eval() method can be used to convert complex numbers as input to the complex objects in Python. To learn more, visit Python eval() function.
Also, notice the way in which the output is formatted. To learn more, visit string formatting in Python.
| true |
e6dead53cca8e5bef774435616a2dd02452e6389 | ejoconnell97/Coding-Exercises | /test2.py | 1,671 | 4.15625 | 4 | def test_2(A):
# write your code in Python 3.6
'''
Loop through S once, saving each value to a new_password
When a numerical character is hit, or end of S, check the new_password
to make sure it has at least one uppercase letter
- Yes, return it
- No, reset string to null and continue through S
If new_password is empty, return -1
'''
current_password = ""
new_password = ""
has_upper = False
password_length = 0
#loop through all of S
for x in A:
#if the current character is not a number
if (x.isalpha()):
current_password += x
#update the new password if needed
new_password, has_upper = update_string(current_password, new_password, has_upper)
else:
#update new_password if S ends with a number
new_password, has_upper = update_string(current_password, new_password, has_upper)
#otherwise, reset the current_password because you've reached a number
current_password = ""
if not new_password:
return -1
print (new_password)
password_length = len(new_password)
return password_length
#make sure that the current_password has at least one uppercase character and is longer than new_password before updating it
def update_string(current, new, has_upper):
for x in current:
if x.isupper():
has_upper = True
if (has_upper == True and (len(current) > len(new))):
new = current
has_upper = False
return new, has_upper
A = "a00000000000000aA9AAA"
test_2(A)
| true |
0860ed14b0d55e8a3c09aeb9e56075354d1e1298 | npandey15/CatenationPython_Assignments | /Task1-Python/Task1_Python/Q3.py | 547 | 4.15625 | 4 | Python 3.8.5 (v3.8.5:580fbb018f, Jul 20 2020, 12:11:27)
[Clang 6.0 (clang-600.0.57)] on darwin
Type "help", "copyright", "credits" or "license()" for more information.
>>> x=6
>>> y=10
>>>
>>> Resultname=x
>>> x=y
>>> y=Resultname
>>> print("The value of x after swapping: {}".format(x))
The value of x after swapping: 10
>>> print("The value of y after swaaping:{}".format(y))
The value of y after swaaping:6
>>>
>>>
>>> # without using third variable
>>>
>>> x=6
>>> y=10
>>>
>>> x,y=y,x
>>> print("x=",x)
x= 10
>>> print("y=",y)
y= 6
>>> | true |
8a99191f6780e06be6e14501b671cc981f42bf0d | npandey15/CatenationPython_Assignments | /Passwd.py | 1,188 | 4.34375 | 4 |
# Passwd creation in python
def checker_password(Passwd):
"Check if Passwd is valid"
Specialchar=["$","#","@","*","%"]
return_val=True
if len(passwd) < 8:
print("Passwd is too short, length of the password should be at least 8")
return_val=False
if len(passwd) > 30:
print("Passwd is too long, length of the password should not be greater than 8")
return_val=False
elif len(passwd) > 8 and len(passwd) <= 30:
print("Passwd OK")
return_val=True
if not any(char.isdigit() for char in passwd):
print("Passwd should have atleast one numerical input")
return_val=False
if not any(char.isupper() for char in passwd):
print("Passwd should have at least one UPPERCASE Letter")
return_val=False
if not any(char in Specialchar for char in passwd):
print("Passwd should have at least one Special character $#@*%")
return_val=False
if return_val:
print("OK")
return return_val
else:
print("Invalid Passwd")
print(checker_password.__doc__)
passwd=input("Enter the password: ")
print(checker_password(passwd))
| true |
6da25701a40c8e6fce75d9ab89aaf63a45e2ecfe | aparna-narasimhan/python_examples | /Strings/shortest_unique_substr.py | 1,322 | 4.15625 | 4 | '''
Smallest Substring of All Characters
Given an array of unique characters arr and a string str, Implement a function getShortestUniqueSubstring that finds the smallest substring of str containing all the characters in arr. Return "" (empty string) if such a substring doesn’t exist.
Come up with an asymptotically optimal solution and analyze the time and space complexities.
Example:
input: arr = ['x','y','z'], str = "xyyzyzyx"
output: "zyx"
See here for more examples:
https://www.geeksforgeeks.org/find-the-smallest-window-in-a-string-containing-all-characters-of-another-string/
TODO: This is a working brute force solution. Need to come up with an optimal solution
'''
from collections import defaultdict
def get_shortest_unique_substring(arr, str):
arr_dict = defaultdict(int)
max_count = len(arr)
min_len = 32768
result = ""
for c in range(len(str)):
length = 0
count = 0
start = 0
for char in arr:
arr_dict[char] = 1
for i,ch in enumerate(str[c:]):
if arr_dict[ch] == 1:
arr_dict[ch] -= 1
count += 1
if start == 0:
start = c
length += 1
if count == max_count:
if length < min_len:
min_len = length
result = str[start:start+length]
#print result
break
return result
| true |
0111fd00103f42a032074e0c3b4d40006f1029f9 | aparna-narasimhan/python_examples | /Arrays/missing_number.py | 866 | 4.28125 | 4 | '''
If elements are in range of 1 to N, then we can find the missing number is a few ways:
Option 1 #: Convert arr to set, for num from 0 to n+1, find and return the number that does not exist in set.
Converting into set is better for lookup than using original list itself. However, space complexity is also O(N).
Option#2: Sum of all elements from 1 to N - Sum of all elements in the array = Missing number
Adv: No space complexity
Option#3: XOR all elements in the array + XOR all elements from 1 to N = Missing number
Provided no duplicates?
Adv: No space complexity
XOR is fast operation
'''
#[1,2,3,5,6,7,8]
def find_missing_number(arr):
n = len(arr)
total_sum = sum(range(1,arr[-1]+1))
print(total_sum)
arr_sum = sum(arr)
print(arr_sum)
missing_num = total_sum - arr_sum
print(missing_num)
find_missing_number([1,2,3,5,6,7,8]) | true |
8d401a11b4ea284f9bb08c2e2615b9414fd3d3ac | aparna-narasimhan/python_examples | /Misc/regex.py | 2,008 | 4.625 | 5 | #https://www.tutorialspoint.com/python/python_reg_expressions.htm
import re
line = "Cats are smarter than dogs";
searchObj = re.search( r'(.*) are (.*?) .*', line, re.M|re.I)
if searchObj:
print "searchObj.group() : ", searchObj.group()
print "searchObj.group(1) : ", searchObj.group(1)
print "searchObj.group(2) : ", searchObj.group(2)
else:
print "Nothing found!!"
'''
When the above code is executed, it produces following result −
searchObj.group() : Cats are smarter than dogs
searchObj.group(1) : Cats
searchObj.group(2) : smarter
'''
#!/usr/bin/python
import re
phone = "2004-959-559 # This is Phone Number"
# Delete Python-style comments
num = re.sub(r'#.*$', "", phone)
print "Phone Num : ", num
# Remove anything other than digits
num = re.sub(r'\D', "", phone)
print "Phone Num : ", num
'''
When the above code is executed, it produces the following result −
Phone Num : 2004-959-559
Phone Num : 2004959559
'''
'''
Sr.No. Example & Description
1
[Pp]ython
Match "Python" or "python"
2
rub[ye]
Match "ruby" or "rube"
3
[aeiou]
Match any one lowercase vowel
4
[0-9]
Match any digit; same as [0123456789]
5
[a-z]
Match any lowercase ASCII letter
6
[A-Z]
Match any uppercase ASCII letter
7
[a-zA-Z0-9]
Match any of the above
8
[^aeiou]
Match anything other than a lowercase vowel
9
[^0-9]
Match anything other than a digit
Special Character Classes
Sr.No. Example & Description
1
.
Match any character except newline
2
\d
Match a digit: [0-9]
3
\D
Match a nondigit: [^0-9]
4
\s
Match a whitespace character: [ \t\r\n\f]
5
\S
Match nonwhitespace: [^ \t\r\n\f]
6
\w
Match a single word character: [A-Za-z0-9_]
7
\W
Match a nonword character: [^A-Za-z0-9_]
Repetition Cases
Sr.No. Example & Description
1
ruby?
Match "rub" or "ruby": the y is optional
2
ruby*
Match "rub" plus 0 or more ys
3
ruby+
Match "rub" plus 1 or more ys
4
\d{3}
Match exactly 3 digits
5
\d{3,}
Match 3 or more digits
6
\d{3,5}
Match 3, 4, or 5 digits
''' | true |
ac26b98728ecf7d49aa79f70aa5dd5e3238ef10d | aparna-narasimhan/python_examples | /pramp_solutions/get_different_number.py | 1,319 | 4.125 | 4 | '''
Getting a Different Number
Given an array arr of unique nonnegative integers, implement a function getDifferentNumber that finds the smallest nonnegative integer that is NOT in the array.
Even if your programming language of choice doesn’t have that restriction (like Python), assume that the maximum value an integer can have is MAX_INT = 2^31-1. So, for instance, the operation MAX_INT + 1 would be undefined in our case.
Your algorithm should be efficient, both from a time and a space complexity perspectives.
Solve first for the case when you’re NOT allowed to modify the input arr. If successful and still have time, see if you can come up with an algorithm with an improved space complexity when modifying arr is allowed. Do so without trading off the time complexity.
Analyze the time and space complexities of your algorithm.
Example:
input: arr = [0, 1, 2, 3]
output: 4
'''
#Approach1: It is enough to iterate only upto length of the list instead of up to MAX_INT. Use set instead of the array itself as lookup time of a set is O(1). Set is similar to dict and can be used when we don't need to store values.
def get_different_number(arr):
temp_set = set(arr)
for i in range(len(arr)):
if i not in temp_set:
return i
if i == (2 ** 31) - 1:
return i
else:
return i + 1 | true |
93354f664979d327f700a7f5ed4a28d92b978b16 | premraval-pr/ds-algo-python | /data-structures/queue.py | 1,040 | 4.34375 | 4 | # FIFO Structure: First In First Out
class Queue:
def __init__(self):
self.queue = []
# Insert the data at the end / O(1)
def enqueue(self, data):
self.queue.append(data)
# remove and return the first item in queue / O(n) Linear time complexity
def dequeue(self):
if self.is_empty():
return -1
data = self.queue[0]
del self.queue[0]
return data
# returns the first element / O(1)
def peek(self):
return self.queue[0]
# check if the queue is empty / O(1)
def is_empty(self):
return self.queue == []
# returns the size / O(1)
def queue_size(self):
return len(self.queue)
queue = Queue()
queue.enqueue(1)
queue.enqueue(2)
queue.enqueue(3)
print("Dequeue: %d" % queue.dequeue())
print("Size: %d" % queue.queue_size())
print("Peek: %d" % queue.peek())
print("Size: %d" % queue.queue_size())
print("Dequeue: %d" % queue.dequeue())
print("Dequeue: %d" % queue.dequeue())
print("Dequeue: %d" % queue.dequeue())
| true |
d73c7a9bec4f30251d31d42a17dfa7fde9a48f8e | danieltapp/fcc-python-solutions | /Basic Algorithm Scripting Challenges/caesars-cipher.py | 848 | 4.21875 | 4 | #One of the simplest and most widely known ciphers is a Caesar cipher, also known as a shift cipher. In a shift cipher the meanings of the letters are shifted by some set amount.
#A common modern use is the ROT13 cipher, where the values of the letters are shifted by 13 places. Thus 'A' ↔ 'N', 'B' ↔ 'O' and so on.
#Write a function which takes a ROT13 encoded string as input and returns a decoded string.
def rot13(str):
l = list(str.lower())
for i in range(0, len(l)):
if l[i].isalpha():
if ord(l[i]) > 109:
l[i] = chr(ord(l[i]) - 13)
elif ord(l[i]) <= 109:
l[i] = chr(ord(l[i]) + 13)
return ''.join(l).upper()
print(rot13("SERR PBQR PNZC"))
print(rot13('SERR CVMMN!'))
print(rot13("SERR YBIR?"))
print(rot13("GUR DHVPX OEBJA QBT WHZCRQ BIRE GUR YNML SBK."))
| true |
2248f6b4e815240d85ee37a6af3dc6c58d44d2d7 | danieltapp/fcc-python-solutions | /Basic Algorithm Scripting Challenges/reverse-a-string.py | 313 | 4.4375 | 4 | #Reverse the provided string.
#You may need to turn the string into an array before you can reverse it.
#Your result must be a string.
def reverse_string(str):
return ''.join(list(reversed(str)))
print(reverse_string('hello'))
print(reverse_string('Howdy'))
print(reverse_string('Greetings from Earth'))
| true |
759f39154fc321a20f84850be8482262b21b427e | oluwaseunolusanya/al-n-ds-py | /chapter_4_basic_data_structures/stackDS.py | 967 | 4.375 | 4 | class Stack:
#Stack implementation as a list.
def __init__(self):
self._items = [] #new stack
def is_empty(self):
return not bool(self._items)
def push(self, item):
self._items.append(item)
def pop(self):
return self._items.pop()
def peek(self):
return self._items[-1]
def size(self):
return len(self._items)
"""
s = Stack()
print(s.is_empty())
s.push('Olu')
print(s.is_empty())
print(s.peek())
print(s.size())
s.push([1 ,2, 5])
print(s.size())
print(s)
print(s.pop())
print(s.size())
"""
#Function below uses Stack to reverse the characters in a string
'''
def rev_string(my_str):
my_str_stack = Stack()
for i in my_str:
my_str_stack.push(i)
print(my_str_stack.peek())
reversed_string = ""
while not my_str_stack.is_empty():
reversed_string += my_str_stack.pop()
return print(reversed_string)
rev_string('I want an apple')
'''
| true |
33c05d544f56bec9699add58b0a26673d41e974a | shubee17/Algorithms | /Array_Searching/search_insert_delete_in_unsort_arr.py | 1,239 | 4.21875 | 4 | # Search,Insert and delete in an unsorted array
import sys
def srh_ins_del(Array,Ele):
# Search
flag = 0
for element in range(len(Array)):
if Array[element] == int(Ele):
flag = 1
print "Element Successfully Found at position =>",element + 1
break
else:
flag = 0
if flag == 0:
print "Element Not Found!!! "
else:
pass
#Insert
print "\n"
flag = 0
position = int(raw_input("Enter the position which you want to be Insert element => "))
Element = int(raw_input("Enter the element => "))
for element in range(len(Array)):
if (element + 1) == int(position):
flag = 1
Array[element] = int(Element)
print "Element Successfully Inserted =>",Array
break
else:
flag = 0
if flag == 0:
print "Position Does'nt Exists!!!"
else:
pass
#Delete
print "\n"
flag = 0
position = int(raw_input("Enter the position which you want to be Delete element => "))
for element in range(len(Array)):
if (element + 1) == int(position):
flag = 1
del(Array[element])
print "Element Successfully Deleted =>",Array
break
else:
flag = 0
if flag == 0:
print "Position Does'nt Exists!!! "
else:
pass
Array = sys.argv[1]
Array = map(int, Array)
Ele = sys.argv[2]
srh_ins_del(Array,Ele)
| true |
b1ec062d544dfb65fbd8f09e67bb03dce65aeef6 | betta-cyber/leetcode | /python/208-implement-trie-prefix-tree.py | 784 | 4.125 | 4 | #!/usr/bin/env python
# encoding: utf-8
class Trie(object):
def __init__(self):
self.root = {}
def insert(self, word):
p = self.root
for c in word:
if c not in p:
p[c] = {}
p = p[c]
p['#'] = True
def search(self, word):
node = self.find(word)
return node is not None and '#' in node
def startsWith(self, prefix):
return self.find(prefix) is not None
def find(self, prefix):
p = self.root
for c in prefix:
if c not in p: return None
p = p[c]
return p
# Your Trie object will be instantiated and called as such:
# obj = Trie()
# obj.insert(word)
# param_2 = obj.search(word)
# param_3 = obj.startsWith(prefix)
| true |
9f7a5977df4e57f7c7c5518e4e3bc42f517d1856 | matthew-lu/School-Projects | /CountingQueue.py | 1,823 | 4.125 | 4 | # This program can be called on to create a queue that stores pairs (x, n) where x is
# an element and n is the count of the number of occurences of x.
# Included are methods to manipulate the Counting Queue.
class CountingQueue(object):
def __init__(self):
self.queue = []
def __repr__(self):
return repr(self.queue)
def add(self, x, count=1):
if len(self.queue) > 0:
xx, cc = self.queue[-1]
if xx == x:
self.queue[-1] = (xx, cc + count)
else:
self.queue.append((x, count))
else:
self.queue = [(x, count)]
def get(self):
if len(self.queue) == 0:
return None
x, c = self.queue[0]
if c == 1:
self.queue.pop(0)
return x
else:
self.queue[0] = (x, c - 1)
return x
def isempty(self):
return len(self.queue) == 0
def countingqueue_len(self):
"""Returns the number of elements in the queue."""
l = self.queue
count = 0
for x, i in l:
for y in range(i):
count += 1
return count
def countingqueue_iter(self):
"""Iterates through all the elements of the queue,
without removing them."""
l = self.queue
for i, x in l:
for y in range(x):
yield i
def subsets(s):
"""Given a set s, yield all the subsets of s,
including s itself and the empty set."""
temp = list(s)
if len(s) == 0:
return [[]]
subset = list(subsets(temp[1:]))
answer = list(subset)
for x in subset:
answer.append(x+[temp[0]])
return answer
return s | true |
8949fd91df9d848465b03c7024e7516738e72c8e | jtew396/InterviewPrep | /linkedlist2.py | 1,651 | 4.21875 | 4 | # Linked List Data Structure
# HackerRank
#
#
# Node class
class Node:
# Function to initialize the node object
def __init__(self, data):
self.data = data
self.next = None
# LinkedList class
class LinkedList:
# Function to initialize the LinkedList class
def __init__(self):
self.head = None
# Function to append a Node object to the LinkedList object
def append(self, data):
if self.head == None:
self.head = Node(data)
return
current = self.head
while current.next != None:
current = current.next
current.next = Node(data)
# Function to prepend a Node object to the LinkedList object
def prepend(self, data):
newHead = Node(data)
newHead.next = self.head
self.head = newHead
# Function to delete a Node object from the LinkedList object
def deleteWithValue(self, data):
if self.head == None:
return
if self.head.data == data:
self.head = self.head.next
return
current = self.head
while current.next != None:
if current.next.data == data:
current.next = current.next.next
return
current = current.next
def printList(self):
temp = self.head
while temp:
print(temp.data)
temp = temp.next
if __name__=='__main__':
llist = LinkedList()
llist.append(1)
llist.append(2)
llist.append(3)
llist.printList()
llist.prepend(0)
llist.printList()
llist.deleteWithValue(0)
llist.printList()
| true |
a624e3d14acd2154c5dc156cfaa092ab1767d6a5 | jtew396/InterviewPrep | /search1.py | 1,154 | 4.25 | 4 | # Cracking the Coding Interview - Search
# Depth-First Search (DFS)
def search(root):
if root == None:
return
print(root)
root.visited = True
for i in root.adjacent:
if i.visited == false:
search(i)
# Breadth-First Search (BFS) - Remember to use a Queue Data Structure
def search(root):
queue = Queue() # Declare a Queue object
root.marked = True # Set the root node marked parameter equal to True
queue.enqueue(root) # Add to the end of the queue
while not queue.isEmpty(): # While the Queue is not empty
r = queue.dequeue() # Remove from the front of the queue
print(r) # Print the returned dequeued node
for i in r.adjacent: # for every node in the adjacent nodes
if i.marked == False: # if they are marked as False
i.marked = True # set them to True
queue.enqueue(i) # Add them to the Queue
# Bidirectional Search Uses 2 Breadth-First Searches
# This is used for to find the shortest path ebetween a source and destination node
| true |
ffbd1535e7f9c25e817c10e31e0e6812a7645724 | knpatil/learning-python | /src/dictionaries.py | 686 | 4.25 | 4 | # dictionary is collection of key-value pairs
# students = { "key1":"value1", "key2": "value2", "key3": "value3" }
# color:point
alien = {} # empty dictionary
alien = {"green":100}
alien['red'] = 200
alien['black'] = 90
print(alien)
# access the value
print(alien['black'])
# modify the value
alien['red'] = 500
print(alien)
# remove the value
# del alien['green']
# print(alien)
# name: favorite programming lanauge
print("__" * 40)
# loop through all key/value pairs
for key,value in alien.items():
print(key, value)
print("__" * 40)
for key in sorted(alien.keys()):
print(key)
print("__" * 40)
for value in alien.values():
print(value)
print("__" * 40)
| true |
7b3f5e98d0ec4f94764e65e70a68c542f44ea966 | knpatil/learning-python | /src/lists2.py | 2,077 | 4.46875 | 4 |
cars = ['bmw', 'audi', 'toyota', 'subaru']
print(cars)
# sort a list by alphabetical order
# cars.sort()
# print(cars)
#
# cars.sort(reverse=True) # reverse order sort
# print(cars)
print(sorted(cars)) # temporarily sort a list
print(cars)
cars.reverse()
print(cars)
print(len(cars))
# print(cars[4]) # exceptions
# comment
for car in cars:
print(car.title()) # 4 spaces of indentation
for value in range(1, 1): # in operator always works with list
print(value)
for number in [1, 2, 4, 2]: #
print(number)
numbers = [11, 232, 44, 0, 2, 5]
smallest_number = min(numbers)
print(smallest_number)
biggest_number = max(numbers)
print(biggest_number)
print("sum of all numbers:", sum(numbers))
# List comprehensions: allows you to generate a list using some code
# list of squares for numbers: 1 - 10
# ** exponent
print("--------")
for x in range(1, 11):
print(x**2)
print("--------")
squares = [x**2 for x in range(1, 11)]
print("squares of 1 - 10: ", squares)
cubes = [x**3 for x in range(1, 11)]
print("cubes of 1 - 10: ", cubes)
# slice of list (part of the list)
print("Cubes of first 5 numbers: ", cubes[0:5]) # 0, 1, 2, 3, 4
print("Cubes of numbers from 2 - 4: ", cubes[1:4]) # 0, 1, 2, 3, 4
start = 0
end = 6
print(cubes[:end]) # slice 0 --> 5 end index non-inclusive
print(cubes[4:]) # slice 0 --> 5 end index non-inclusive
print(cubes[:])
# slice of last 3 numbers in the list
print(cubes[-3:])
print(cubes[-3:-1]) #
for number in cubes[4:8]: # list slice from 4th to 7th item
print(number)
squares2 = squares
print(squares)
squares[0] = 9999
print("squares:", squares)
print("squares2:", squares2)
squares3 = squares[:] # new slice
print(squares3)
squares[0] = 1111
print(squares3)
squares[5] = 42942940
print(squares)
# lists in python are mutable(changeable): that means you can change/modify the values
# tuple -- immutable <-- you cannot change values
x = (2, 4) # using a round bracket, it is a tuple
print(x)
print(x[0]) # accessing the value
for i in x:
print(i)
x = (3, 9) # types are dynamic
| true |
a87bf15fb1e4f742f5f31be7a1af84276ffe6fc2 | MaxAttax/maxattax.github.io | /resources/Day1/00 - Python Programming/task4.py | 808 | 4.40625 | 4 | # Task 4: Accept comma-separated input from user and put result into a list and into a tuple
# The program waits until the user inputs a String into the console
# For this task, the user should write some comma-separated integer values into the console and press "Enter"
values = input() # Use for Python 3
# After the user pressed "Enter", the rest of the program from here is executed
# The split()-operator splits a given String at the indicated separator (here a comma ",")
# The split(",") function is used with the value that carries the user input from above (values)
l = values.split(",")
# A tuple is a special kind of list that can not be modified or extended.
# The tuple() function generates a tuple from a given list.
t = tuple(l)
# Finally the results are printed out.
print(l)
print(t) | true |
e829a743405b49893446c705272db2f7323bb329 | Sharanhiremath02/C-98-File-Functions | /counting_words_from_file.py | 281 | 4.375 | 4 | def count_words_from_file():
fileName=input("Enter the File Name:")
file=open(fileName,"r")
num_words=0
for line in file:
words=line.split()
num_words=num_words+len(words)
print("number of words:",num_words)
count_words_from_file() | true |
7c821b9a5102495f8f22fc21e29dd9812c77b3bd | chetanDN/Python | /AlgorithmsAndPrograms/02_RecursionAndBacktracking/01_FactorialOfPositiveInteger.py | 222 | 4.125 | 4 | #calculate the factorial of a positive integer
def factorial(n):
if n == 0: #base condition
return 1
else:
return n * factorial(n-1) #cursive condition
print(factorial(6))
| true |
9bbcd87d994a200ebe11bc09ff79995dd74eac0a | GBaileyMcEwan/python | /src/hello.py | 1,863 | 4.5625 | 5 | #!/usr/bin/python3
print("Hello World!\n\n\n");
#print a multi-line string
print(
'''
My
Multi
Line
String
'''
);
#concatinate 2 words
print("Pass"+"word");
#print 'Ha' 4 times
print("Ha" * 4);
#get the index of the letter 'd' in the word 'double'
print("double".find('d'));
#print a lower-case version of the string
print("This Sucks".lower());
#print special characters- double escape if you want to print a backslash
print("This rocks\tThis rocks\\t");
#If you want to use double quotes here, you would need to escape them
print("Single 'quotes' are fine but \"doubles\" are a problem");
#You can use / * + - as well as // (returns whole numbers only and ** which returns exponents and % which mods
print(2/2);
#variables are like the below
my_string = "My sample string";
my_string += " string2";
my_int = 2+2;
print(my_string + " testing ");
print(my_int);
#this is how you print a list/array
my_list = [1,2,"three",True];
print(my_list[1]);
#print the length of the list
print(len(my_list));
#s is how you print from the index position and how many items in the list you want to printt
print(my_list[3:len(my_list)]);
#You can add items to the list as per below
my_list.append(5);
print(my_list[0:]);
#you can also remove items from the list:
my_list.remove("three");
print(my_list);
print("My string is " + str(my_int) + " what");
#hashes or dictionaries are set like this
ages = {'kevin': 59, 'alex': 60}
print(ages);
print(ages['kevin']);
ages['kayla']=21;
print(ages);
ages['kayla']=22;
print(ages);
del ages['kayla'];
print(ages);
#use this as a safer way to delete an element from a hash
ages.pop('alex');
print(ages);
#print only the keys or values in the hash
print(ages.keys());
print(ages.values());
print(list(ages.values()));
#you can also create a hash/dictionary like this
weights = dict(kevin=100, bob=200);
print(weights);
| true |
34ad45e897f4b10154ccf0f0585c29e1e51ad2f8 | EwarJames/alx-higher_level_programming | /0x0B-python-input_output/3-to_json_string.py | 315 | 4.125 | 4 | #!/usr/bin/python3
"""Define a function that returns a json."""
import json
def to_json_string(my_obj):
"""
Function that convert json to a string.
Args:
my_obj (str): Object to be converted.
Return:
JSON representation of an object (string)
"""
return json.dumps(my_obj)
| true |
431384abd81af78ff79355d261528645cf9c100b | pi408637535/Algorithm | /com/study/algorithm/daily/diameter-of-binary-tree.py | 1,370 | 4.15625 | 4 | '''
树基本是递归。
递归：四要素
本题思路:left+right=diameter
https://www.lintcode.com/problem/diameter-of-binary-tree/description
'''
#思路：
#Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
class Solution:
"""
@param root: a root of binary tree
@return: return a integer
"""
def helper(self, root):
if not root:
return 0
left_depth = self.helper(root.left)
right_depth = self.helper(root.right)
depth = max(left_depth, right_depth) + 1
return depth
def diameterOfBinaryTree(self, root):
# write your code here
left_depth = self.helper(root.left)
right_depth = self.helper(root.right)
return left_depth + right_depth
if __name__ == '__main__':
TreeNode1 = TreeNode(1)
TreeNode2 = TreeNode(2)
TreeNode3 = TreeNode(3)
TreeNode4 = TreeNode(4)
TreeNode5 = TreeNode(5)
TreeNode1.left = TreeNode2
TreeNode1.right = TreeNode3
TreeNode2.left = TreeNode4
TreeNode2.right = TreeNode5
# TreeNode1 = TreeNode(1)
# TreeNode2 = TreeNode(2)
# TreeNode3 = TreeNode(3)
#
# TreeNode2.left = TreeNode3
# TreeNode3.left = TreeNode1
print(Solution().diameterOfBinaryTree(TreeNode1))
| true |
8661d933bf731b111cda9345752a0307ff4adca8 | rodrigomanhaes/model_mommy | /model_mommy/generators.py | 1,794 | 4.34375 | 4 | # -*- coding:utf-8 -*-
__doc__ = """
Generators are callables that return a value used to populate a field.
If this callable has a `required` attribute (a list, mostly), for each item in the list,
if the item is a string, the field attribute with the same name will be fetched from the field
and used as argument for the generator. If it is a callable (which will receive `field`
as first argument), it should return a list in the format (key, value) where key is
the argument name for generator and value is the value for that argument.
"""
import sys
import string
import datetime
from decimal import Decimal
from random import randint, choice, random
MAX_LENGTH = 300
def gen_from_list(L):
'''Makes sure all values of the field are generated from the list L
Usage:
from mommy import Mommy
class KidMommy(Mommy):
attr_mapping = {'some_field':gen_from_list([A, B, C])}
'''
return lambda: choice(L)
# -- DEFAULT GENERATORS --
def gen_from_choices(C):
choice_list = map(lambda x:x[0], C)
return gen_from_list(choice_list)
def gen_integer(min_int=-sys.maxint, max_int=sys.maxint):
return randint(min_int, max_int)
gen_float = lambda:random()*gen_integer()
def gen_decimal(max_digits, decimal_places):
num_as_str = lambda x: ''.join([str(randint(0,9)) for i in range(x)])
return "%s.%s" % (
num_as_str(max_digits-decimal_places),
num_as_str(decimal_places)
)
gen_decimal.required = ['max_digits', 'decimal_places']
gen_date = datetime.date.today
gen_datetime = datetime.datetime.now
def gen_string(max_length):
return ''.join(choice(string.printable) for i in range(max_length))
gen_string.required = ['max_length']
gen_text = lambda: gen_string(MAX_LENGTH)
gen_boolean = lambda: choice((True, False))
| true |
6c036ab3ae9e679019d589fbd8de8be45486773f | gbrough/python-projects | /palindrome-checker.py | 562 | 4.375 | 4 | # Ask user for input string
# Reverse the string
# compare if string is equal
# challenge - use functions
word = None
def wordInput():
word = input("Please type a word you would like to see if it's a palindrome\n").lower()
return word
def reverseWord():
reversedWord = word[::-1]
return reversedWord
def palindromeCheck():
if word == reverseWord:
print("Yes, you have entered a palindrome")
else:
print("Sorry, this word is not a palindrome")
def main():
print("\nPalindrome Checker")
wordInput
reverseWord
palindromeCheck
main() | true |
5ebe1f88e466c99ba52f3dd43a17e692b30c96b2 | chena/aoc-2017 | /day12.py | 2,625 | 4.21875 | 4 | """
part 1:
Each program has one or more programs with which it can communicate, and they are bidirectional;
if 8 says it can communicate with 11, then 11 will say it can communicate with 8.
You need to figure out how many programs are in the group that contains program ID 0.
For example, suppose you go door-to-door like a travelling salesman and record the following list:
0 <-> 2
1 <-> 1
2 <-> 0, 3, 4
3 <-> 2, 4
4 <-> 2, 3, 6
5 <-> 6
6 <-> 4, 5
In this example, the following programs are in the group that contains program ID 0:
- Program 0 by definition.
- Program 2, directly connected to program 0.
- Program 3 via program 2.
- Program 4 via program 2.
- Program 5 via programs 6, then 4, then 2.
- Program 6 via programs 4, then 2.
Therefore, a total of 6 programs are in this group; all but program 1, which has a pipe that connects it to itself.
How many programs are in the group that contains program ID 0?
"""
def __check_connections__(program_id, connections, group):
for c in connections[program_id]:
if c in group:
continue
group.add(c)
__check_connections__(c, connections, group)
def __get_connetions__():
with open('input/programs.txt') as f:
content = f.readlines()
connections = {}
programs = [p.strip() for p in content]
arrow = '<->'
for p in programs:
parts = p.split(arrow)
program = int(parts[0].strip())
connections[program] = [int(l) for l in parts[1].strip().split(', ')]
return connections
def digital_plumber_zero_group():
connections = __get_connetions__()
# traverse through each program and check its connections
# keep track of the programs with a set
zero_group = {0}
__check_connections__(0, connections, zero_group)
return len(zero_group)
# print(digital_plumber_zero_group())
"""
part 2:
A group is a collection of programs that can all communicate via pipes either directly or indirectly.
The programs you identified just a moment ago are all part of the same group.
Now, they would like you to determine the total number of groups.
In the example above, there were 2 groups:
one consisting of programs 0,2,3,4,5,6, and the other consisting solely of program 1.
How many groups are there in total?
"""
def digital_plumber_group_count():
connections = __get_connetions__()
# traverse through each program mark the group it belongs to
groupings = {}
for p in connections:
if p in groupings:
continue
group = {p}
__check_connections__(p, connections, group)
for member in group:
groupings[member] = p
return len(set(groupings.values()))
print(digital_plumber_group_count()) | true |
6e33fd4e81dcfdce22916bd20d4230e472490dae | chena/aoc-2017 | /day05.py | 1,882 | 4.53125 | 5 | """
part 1:
The message includes a list of the offsets for each jump.
Jumps are relative: -1 moves to the previous instruction, and 2 skips the next one.
Start at the first instruction in the list. The goal is to follow the jumps until one leads outside the list.
In addition, these instructions are a little strange; after each jump, the offset of that instruction increases by 1.
So, if you come across an offset of 3, you would move three instructions forward, but change it to a 4 for the next time it is encountered.
(0) 3 0 1 -3 - before we have taken any steps.
(1) 3 0 1 -3 - jump with offset 0 (that is, don't jump at all). Fortunately, the instruction is then incremented to 1.
2 (3) 0 1 -3 - step forward because of the instruction we just modified. The first instruction is incremented again, now to 2.
2 4 0 1 (-3) - jump all the way to the end; leave a 4 behind.
2 (4) 0 1 -2 - go back to where we just were; increment -3 to -2.
2 5 0 1 -2 - jump 4 steps forward, escaping the maze.
In this example, the exit is reached in 5 steps.
part 2:
After each jump, if the offset was three or more, instead decrease it by 1.
Otherwise, increase it by 1 as before.
Using this rule with the above example, the process now takes 10 steps,
and the offset values after finding the exit are left as 2 3 2 3 -1.
How many steps does it now take to reach the exit?
"""
def __parse_maze_input__(filename):
with open(filename) as f:
content = f.readlines()
return [int(n) for n in content]
def maze_of_twisty(maze):
steps = 0
index = 0
while (index < len(maze)):
move_forward = maze[index]
if (move_forward >= 3):
maze[index] -= 1
else:
maze[index] += 1
index += move_forward
steps += 1
return steps
print(maze_of_twisty([0, 3, 0, 1, -3]))
# print(maze_of_twisty(__parse_maze_input__('input/maze.txt'))) | true |
164dda3ecb9d27c153dbc9d143ba05437c47f656 | luisprooc/data-structures-python | /src/bst.py | 1,878 | 4.1875 | 4 | from binary_tree import Node
class BST(object):
def __init__(self, root):
self.root = Node(root)
def insert(self, new_val):
current = self.root
while current:
if new_val >= current.value:
if not current.right:
current.right = Node(new_val)
return
current = current.right
else:
if not current.left:
current.left = Node(new_val)
return
current = current.left
def search(self, find_val):
# If current node is equal to
# find_value, return True
if self.root.value == find_val:
return True
else:
# If the find_value is bigger to current node and the current
# node not is a leaf, scroll right
if find_val >= self.root.value and self.root.right:
self.root = self.root.right
return self.search(find_val)
else:
# If current node not is a leaf, scroll left
if self.root.left:
self.root = self.root.right
return self.search(find_val)
# This means that you reached a leaf and not
# find the expected value
return False
def preorder_print(self, start, traversal):
if start:
traversal += (str(start.value) + "-")
traversal = self.preorder_print(start.left, traversal)
traversal = self.preorder_print(start.right, traversal)
return traversal
# Set up tree
tree = BST(4)
# Insert elements
tree.insert(2)
tree.insert(1)
tree.insert(3)
tree.insert(5)
# Check search
# Should be True
print (tree.search(4))
# Should be False
print (tree.search(6)) | true |
757cfb65fc22c1f8f9326a25018defb7aafbad56 | 2FLing/CSCI-26 | /codewar/camel_case.py | 531 | 4.25 | 4 | # Complete the method/function so that it converts dash/underscore delimited words into camel casing. The first word within the output should be capitalized only if the original word was capitalized (known as Upper Camel Case, also often referred to as Pascal case).
# Examples
# to_camel_case("the-stealth-warrior") # returns "theStealthWarrior"
# to_camel_case("The_Stealth_Warrior") # returns "TheStealthWarrior"
def to_camel_case(s):
return s[0]+s.translate(str.maketrans("","","-_"))[1:]
print(to_camel_case("I_love-You")) | true |
5a5581ad8604d9d61d579ab1661be5c7d70cdcb6 | rodrigocamarena/Homeworks | /sess3&4ex4.py | 2,915 | 4.5 | 4 | print("Welcome to the universal pocket calculator.")
print("1. Type <+> if you want to compute a sum."
"\n2. Type <-> if you want to compute a rest."
"\n3. Type <*> if you want to compute a multiplication."
"\n4. Type </> if you want to compute a division."
"\n5. Type <quit> if you want to exit.")
while True:
prompt = 'Introduce your answer: '
answer = input(prompt)
if answer == "+":
while True:
try:
num = input('Introduce the first number: ')
num2 = input('Introduce the second number: ')
num = float(num)
num2 = float(num2)
except ValueError:
print("opps, please give me a proper number!")
except:
print("opps, something unknown occured.")
else:
result = num + num2
print("The result of: ",num,"+",num2," is: ",result)
break
elif answer == "-":
while True:
try:
num = input('Introduce the first number: ')
num2 = input('Introduce the second number: ')
num = float(num)
num2 = float(num2)
except ValueError:
print("opps, please give me a proper number!")
except:
print("opps, something unknown occured.")
else:
result = num - num2
print("The result of: ",num,"-",num2," is: ",result)
break
elif answer == "*":
while True:
try:
num = input('Introduce the first number: ')
num2 = input('Introduce the second number: ')
num = float(num)
num2 = float(num2)
except ValueError:
print("opps, please give me a proper number!")
except:
print("opps, something unknown occured.")
else:
result = num * num2
print("The result of: ",num,"*",num2," is: ",result)
break
elif answer == "/":
while True:
try:
num = input('Introduce the first number: ')
num2 = input('Introduce the second number: ')
num = float(num)
num2 = float(num2)
except ValueError:
print("opps, please give me a proper number!")
except:
print("opps, something unknown occured.")
else:
result = num / num2
print("The result of: ", num, "/", num2, " is: ", result)
break
elif answer == "quit":
print("Was a pleasure working with you. See you soon!")
break
else:
print("opps, invalid answer. Try again please!")
| true |
7536d2a8c16cc9bda41d4a4d3894c0a8a0911446 | artemis-beta/phys-units | /phys_units/examples/example_2.py | 1,160 | 4.34375 | 4 | ##############################################################################
## Playing with Measurements ##
## ##
## In this example the various included units are explored in a fun and ##
## silly way to illustrate how units_database behaves. ##
## ##
##############################################################################
#--------------------- Fetch all the units we need --------------------------#
from units_database import mile, m, furlong, yd, pc, rod
my_distance = 5*mile
# By default units_database will use SI units the function 'as_base' treats
# the unit as a base unit.
print('Today I ran {}.'.format(my_distance.as_base()))
print('This is equivalent to {}'.format(my_distance))
print('or (if we want to be silly), this is also {}'.format(my_distance.as_unit(pc)))
print('In Imperial units this is also {}'.format(my_distance.as_unit(yd)))
print('or (to be unusual) {}'.format(my_distance.as_unit(furlong))) | true |
3c419f360e36ed45af7d7935aa2f824bb2dc472a | Cleancode404/ABSP | /Chapter8/input_validation.py | 317 | 4.1875 | 4 | import pyinputplus as pyip
while True:
print("Enter your age:")
age = input()
try:
age = int(age)
except:
print("Please use numeric digits.")
continue
if age < 0:
print("Please enter a positive number.")
continue
break
print('Your age is', age) | true |
bf005c98b3c7beabc0e2000cfd0cfd404010a9a9 | AlexRoosWork/PythonScripts | /delete_txts.py | 767 | 4.15625 | 4 | #!/usr/bin/env python3
# given a directory, go through it and its nested dirs to delete all .txt files
import os
def main():
print(f"Delete all text files in the given directory\n")
path = input("Input the basedir:\n")
to_be_deleted = []
for dirpath, dirnames, filenames in os.walk(path):
for filename in filenames:
if filename.endswith(".txt"):
string = dirpath + os.sep + filename
to_be_deleted.append(string)
num_of_files = len(to_be_deleted)
if input(f"sure you want to delete {num_of_files} .txt files? (y/n)") == "y":
for file in to_be_deleted:
os.remove(file)
print("Done.")
else:
print("Abortion.")
if __name__ == "__main__":
main()
| true |
8ae3f309e572b41a3ff5205692f0ae4c90f11962 | Trenchevski/internship | /python/ex6.py | 867 | 4.21875 | 4 | # Defining x with string value
x = "There are %d types of people." % 10
# Binary variable gets string value with same name
binary = "binary"
# Putting string value to "don't"
do_not = "don't"
# Defining value of y variable using formatters
y = "Those who know %s and those who %s." % (binary, do_not)
# Printing value of x
print x
# Printing value of y
print y
# Printing value of x using formatters
print "I said: %r." % x
# Printing value of y using formatters
print "I also said: '%s'." % y
# Boolean with value False
hilarious = False
# Variable joke_evaluation gets string value
joke_evaluation = "Isn't that joke so funny?! %r"
# Printing mod of two previous variables
print joke_evaluation % hilarious
# Putting string value to w and e variables
w = "This is the left side of..."
e = "a string with a right side."
# Printing them one after another
print w + e
| true |
6cd204d47bb1937a024c1afa0c25527316453468 | Soares/natesoares.com | /overviewer/utilities/string.py | 633 | 4.5 | 4 | def truncate(string, length, suffix='...'):
"""
Truncates a string down to at most @length characters.
>>> truncate('hello', 12)
'hello'
If the string is longer than @length, it will cut the
string and append @suffix to the end, such that the
length of the resulting string is @length.
>>> truncate('goodbye', 4)
'g...'
>>> truncate('goodbye', 6)
'goo...'
@suffix defaults to '...'.
>>> truncate('hello', 3, '..')
'h..'
>>> truncate('hello', 3, '')
'hel'
"""
if len(string) <= length:
return string
return string[:length-len(suffix)] + suffix
| true |
066bcfb00c4f01528d79d8a810a65d2b64e8a8a2 | bchaplin1/homework | /week02/03_python_homework_chipotle.py | 2,812 | 4.125 | 4 | '''
Python Homework with Chipotle data
https://github.com/TheUpshot/chipotle
'''
'''
BASIC LEVEL
PART 1: Read in the data with csv.reader() and store it in a list of lists called 'data'.
Hint: This is a TSV file, and csv.reader() needs to be told how to handle it.
https://docs.python.org/2/library/csv.html
'''
import csv
cd 'c:\ml\dat7\data'
with open ('chipotle.tsv','rU') as f:
data = [row for row in csv.reader(f,delimiter='\t')]
'''
BASIC LEVEL
PART 2: Separate the header and data into two different lists.
'''
header = data[0]
data = data[1:]
'''
INTERMEDIATE LEVEL
PART 3: Calculate the average price of an order.
Hint: Examine the data to see if the 'quantity' column is relevant to this calculation.
Hint: Think carefully about the simplest way to do this!
'''
sum([float(d[4].replace('$','')) for d in data])/len(set([d[0] for d in data]))
'''
INTERMEDIATE LEVEL
PART 4: Create a list (or set) of all unique sodas and soft drinks that they sell.
Note: Just look for 'Canned Soda' and 'Canned Soft Drink', and ignore other drinks like 'Izze'.
'''
drinks = set()
for d in data:
if d[2] == 'Canned Soda' or d[2] == 'Canned Soft Drink':
drinks.add(d[3])
print drinks
'''
ADVANCED LEVEL
PART 5: Calculate the average number of toppings per burrito.
Note: Let's ignore the 'quantity' column to simplify this task.
Hint: Think carefully about the easiest way to count the number of toppings!
'''
topCt = 0
burritoCt = 0
for d in data:
if d[2] == 'Burrito' :
topCt += len(d[3].split('[')[-1].split(','))
burritoCt+=1
avgCt = topCt / float(burritoCt)
print 'average number of toppings per burrito is ' + str(avgCt)
'''
ADVANCED LEVEL
PART 6: Create a dictionary in which the keys represent chip orders and
the values represent the total number of orders.
Expected output: {'Chips and Roasted Chili-Corn Salsa': 18, ... }
Note: Please take the 'quantity' column into account!
Optional: Learn how to use 'defaultdict' to simplify your code.
'''
from collections import defaultdict
chips = [d for d in data if d[2].startswith('Chips')]
d = defaultdict(int)
for order in chips:
d[order[2]]+=int(order[1])
print d.items()
'''
BONUS: Think of a question about this data that interests you, and then answer it!
'''
'''
is there a correlation between meat selection and toppings, like fat content?
let's put them in a dictionary and find out
'''
chicken = [d for d in data if d[2].startswith('Chicken')]
steak = [d for d in data if d[2].startswith('Steak')]
orderSum = 0
print 'chicken' + str(sum([float(d[4].replace('$','')) for d in chicken])/len(set([d[0] for d in chicken])))
print 'steak' + str(sum([float(d[4].replace('$','')) for d in steak])/len(set([d[0] for d in steak])))
| true |
90af163267b8d485c28169c9aeb149df021e3509 | hemenez/holbertonschool-higher_level_programming | /0x07-python-test_driven_development/0-add_integer.py | 423 | 4.3125 | 4 | #!/usr/bin/python3
def add_integer(a, b):
"""Module will add two integers
"""
total = 0
if type(a) is not int and type(a) is not float:
raise TypeError('a must be an integer')
if type(b) is not int and type(b) is not float:
raise TypeError('b must be an integer')
if type(a) is float:
a = int(a)
if type(b) is float:
b = int(b)
total = a + b
return total
| true |
ba2c8e1e768b24f7ad7a4c00ee6ed4116d31a21a | kjigoe/Earlier-works | /Early Python/Fibonacci trick.py | 866 | 4.15625 | 4 | dic = {0:0, 1:1}
def main():
n = int(input("Input a number"))
## PSSST! If you use 45 for 'n' you get a real phone number!
counter = Counter()
x = fib(n,counter)
print("Fibonacci'd with memoization I'd get",x)
print("I had to count",counter,"times!")
y = recursivefib(n, counter)
print("And with recusion I still get",y)
print("But the count changes to",counter)
def fib(n,counter):
if n in dic:
return dic[n]
else:
counter.increment()
if n < 2:
dic[n] = n
else:
dic[n] = fib(n-2,counter) + fib(n-1,counter)
return dic[n]
def recursivefib(n, counter):
if n < 2:
return n
else:
counter.increment()
return (recursivefib(n-2,counter) + recursivefib(n-1,counter))
class Counter(object):
def __init__(self):
self._number = 0
def increment(self):
self._number += 1
def __str__(self):
return str(self._number)
main()
| true |
e996c9bff605c46d30331d13e44a49c04a2e29be | Kaylotura/-codeguild | /practice/greeting.py | 432 | 4.15625 | 4 | """Asks for user's name and age, and greets them and tells them how old they'll be next year"""
# 1. Setup
# N/A
# 2. Input
name = input ("Hello, my name is Greetbot, what's your name? ")
age = input(name + ' is a lovely name! How old are you, ' + name + '? ')
# 3. Transform
olderage = str(int(age) + 1)
# 4. Output
print ("You're " + age + "-years old? Wow, I guess that means you're going to be " + olderage + " next year!")
| true |
922c0d74cf538e3a28a04581b9f57f7cfb7377e4 | BstRdi/wof | /wof.py | 1,573 | 4.25 | 4 | from random import choice
"""A class that can be used to represent a wheel of fortune."""
fields = ('FAIL!', 'FAIL!', 100, 'FAIL!', 'FAIL!', 500, 'FAIL!', 250, 'FAIL!', 'FAIL!', 'FAIL!', 'FAIL!', 1000, 'FAIL!', 'FAIL!', 'FAIL!', 'FAIL!', 'FAIL!', 'FAIL!')
score = []
class WheelOfFortune:
"""A simple attempt to represent a wheel of fortune."""
def __init__(self, fields, spins=10):
"""Initialize attributes to describe a wheel of fortune."""
self.fields = fields
self.spins = spins
def spin_wheel(self):
"""Spin the wheel of fortune"""
for spin in range(1, self.spins):
spin = choice(fields)
print(spin)
if isinstance(spin, int) == True:
score.append(spin)
sum_score = sum(score)
print(f'{player_name} you got {sum_score} points!')
# Spin for player one
player_name = input('Player One - Enter your name: ')
player_one = WheelOfFortune(fields)
player_one.spin_wheel()
player_one_score = score[:]
player_one_name = player_name
del score[:]
# Spin for player two
player_name = input('Player Two - Enter your name: ')
player_two = WheelOfFortune(fields)
player_two.spin_wheel()
player_two_score = score[:]
player_two_name = player_name
# Selection of the winner
if player_one_score > player_two_score:
print(f'{player_one_name} wins!')
elif player_two_score > player_one_score:
print(f'{player_two_name} wins!')
elif player_one_score == player_two_score:
print('Draw!') | true |
b63879f6a16ae903c1109d3566089e47d0212200 | idahopotato1/learn-python | /01-Basics/005-Dictionaries/dictionaries.py | 1,260 | 4.375 | 4 | # Dictionaries
# A dictionary is an associative array (also known as hashes).
# Any key of the dictionary is associated (or mapped) to a value.
# The values of a dictionary can be any Python data type.
# So dictionaries are unordered key-value-pairs.
# Constructing a Dictionary
my_dist = {'key1': 'value1', 'key2': 100, 'key3': 99.99}
# Accessing Objects from a dictionary
print(my_dist['key2']) # 100
print(my_dist['key1'].upper()) # VALUE1
print(my_dist['key2'] - 2) # 98
print(my_dist['key2']) # 100
my_dist['key2'] = my_dist['key2'] - 2
print(my_dist['key2']) # 98
print(my_dist) # {'key1': 'value1', 'key2': 98, 'key3': 99.99}
my_colors = {}
my_colors['color1'] = 'red'
my_colors['color2'] = 'green'
print(my_colors) # {'color1': 'red', 'color2': 'green'}
# Nesting dictionary
my_dist2 = {'key1': {'subKey': 'value'}, 'key2': 100, 'key3': 99.99}
print(my_dist2['key1']['subKey'].upper()) # VALUE
# Basic Dictionary Methods
my_dist3 = {'key1': 'value1', 'key2': 'value2', 'key3': 'value3'}
print(my_dist3.keys()) # dict_keys(['key1', 'key2', 'key3'])
print(my_dist3.values()) # dict_values(['value1', 'value2', 'value3'])
print(my_dist3.items()) # dict_items([('key1', 'value1'), ('key2', 'value2'), ('key3', 'value3')])
| true |
b5dbb2bc21aca13d23b3d3f87569877ce9951eec | idahopotato1/learn-python | /04-Methods-Functions/001-methods.py | 736 | 4.34375 | 4 | # Methods
# The other kind of instance attribute reference is a method.
# A method is a function that “belongs to” an object.
# (In Python, the term method is not unique to class instances:
# other object types can have methods as well.
# For example, list objects have methods called append, insert, remove, sort, and so on.
print('=============================================')
my_list = [1, 2, 3]
my_list.append(4)
print(my_list) # [1, 2, 3, 4]
print('=============================================')
print(my_list.count(3)) # 1
print('=============================================')
print(help(my_list.append)) # append(...) method of builtins.list instance
print('=============================================')
| true |
4afb88e53ccddb16c631b2af181bb0e607a2b37b | Evakung-github/Others | /381. Insert Delete GetRandom O(1).py | 2,170 | 4.15625 | 4 | '''
A hashmap and an array are created. Hashmap tracks the position of value in the array, and we can also use array to track the appearance in the hashmap.
The main trick is to swap the last element and the element need to be removed, and then we can delete the last element at O(1) cost.
Afterwards, we need to update the position of the original last element in the hashmap to its current position.
Therefore, that's why we need to record its space in the value of the hashmap.
ref: https://www.youtube.com/watch?v=mRTgft9sBhA
'''
import random
class RandomizedCollection:
def __init__(self):
"""
Initialize your data structure here.
"""
self.array = []
self.dict = {}
def insert(self, val: int) -> bool:
"""
Inserts a value to the collection. Returns true if the collection did not already contain the specified element.
"""
if val not in self.dict:
self.dict[val] = [len(self.array)]
self.array.append([val,0])
return True
else:
self.dict[val].append(len(self.array))
self.array.append([val,len(self.dict[val])-1])
def remove(self, val: int) -> bool:
"""
Removes a value from the collection. Returns true if the collection contained the specified element.
"""
if val in self.dict:
if not self.dict[val]:
return False
else:
n = self.dict[val][-1]
self.dict[self.array[-1][0]][self.array[-1][1]] = n
n = self.dict[val].pop()
self.array[n], self.array[-1] = self.array[-1],self.array[n]
self.array.pop()
return True
def getRandom(self) -> int:
"""
Get a random element from the collection.
"""
n = random.randint(0,len(self.array)-1)
return self.array[n][0]
# Your RandomizedCollection object will be instantiated and called as such:
# obj = RandomizedCollection()
# param_1 = obj.insert(val)
# param_2 = obj.remove(val)
# param_3 = obj.getRandom()
| true |
c57eab0302c15814a5f51c2cbc0fa104910eef08 | hihihien/nrw-intro-to-python | /lecture-06/solutions/exponent.py | 261 | 4.28125 | 4 | base = float(input('What is your base?'))
exp = float(input('What is your exponent?'))
num = exp
result = 1
while num > 0:
result = result * base
num = num - 1
print(f'{base} raised to the power of {exp} is: {result}. ({base} ** {exp} = {base**exp})')
| true |
ef2e1747c49dca4e17fe558704d05d50b2a11506 | kengbailey/interview-prep | /selectionsort.py | 1,507 | 4.28125 | 4 | # Selection Sort Implementation in Python
'''
How does it work?
for i = 1:n,
k = i
for j = i+1:n, if a[j] < a[k], k = j
→ invariant: a[k] smallest of a[i..n]
swap a[i,k]
→ invariant: a[1..i] in final position
end
What is selection sort?
The selection sort algorithm is a combination of searching and sorting.
It sorts an array by repeatedly finding the minimum/maximum element from unsorted part
and putting it at the end.
In selection sort, the inner loop finds the next smallest (or largest) value and the outer
loop places that value into its proper location.
Selection sort should never be used. it does not adapt to the data in any way.
What is performance of selection sort in Big'O?
Worst-case performance О(n2)
Best-case performance О(n2)
Average performance О(n2)
Worst-case space Complexity О(n) total, O(1) auxiliary
'''
def selection_sort(unsorted_arr):
for x in range(len(unsorted_arr)):
# store current position
n = x
# search for the position of the next smallest value
# in the unsorted part
for y in range(x+1, len(unsorted_arr)):
if unsorted_arr[y] < unsorted_arr[n]:
n = y
# swap next smallest value into postition
swap(x,n,unsorted_arr)
print(unsorted_arr)
def swap(a, b, array):
temp2 = array[a]
array[a] = array[b]
array[b] = temp2
arr = [10,9,7,8,6,5,4,2,3,1]
print("Unsorted: ", arr)
selection_sort(arr)
print("Sorted: ", arr) | true |
6c612b3a3904a9710b3f47c0174edf1e0f15545b | spots1000/Python_Scripts | /Zip File Searcher.py | 2,412 | 4.15625 | 4 | from zipfile import ZipFile
import sys
import os
#Variables
textPath = "in.txt"
outPath = "out.txt"
## Announcements
print("Welcome to the Zip Finder Program!")
print("This program will take a supplied zip file and locate within said file any single item matching the strings placed in an accompanying text file.")
input("To use this program please place the desired skus or other identifies in a folder called \"" + textPath + "\" and then press Enter to continue.\n")
## Ensure before opening text file that we have a valid file to read.
if not (os.path.isfile(textPath)):
print("A valid text file was not found, program will now exit.")
input("Press any key to exit...")
sys.exit()
## Attempt to open our text file and read in the data
targetList = []
textFile = open(textPath, encoding="utf8")
for line in textFile:
strippedLine = line.strip("\n")
strippedLine = strippedLine.strip("\t")
targetList.append(strippedLine)
print("Text File successfully read in with " + str(len(targetList)) + " items.")
textFile.close()
## Get the filename
fileName = input("Please enter the name of the zip file without any extensions. >>>")
fileName = fileName + ".zip"
## Make sure the file exists
try:
zip = ZipFile(fileName, 'r')
except:
print("Zip File Was Not Opened Successfully")
print("Program will exit")
sys.exit()
print("Zip file was read successfully, processing will now comence")
## Read the zip file
fileData = zip.infolist()
errorList = []
## Loop through the data for the relevant
for target in targetList:
print("Attemtping to locate " + target)
found = "false"
for dat in fileData:
if (target in dat.filename):
print("Located target file")
path = zip.extract(dat)
found = "true"
break;
if(found == "false"):
print("Target was not found.")
errorList.append(target)
## output to the output file
outFile = open(outPath, "a")
for err in errorList:
outFile.write(err + "\n")
outFile.close()
i = len(targetList) - len(errorList)
##Final output
print("All processing completed. " + str(i) + " total folders successfully processed while " + str(len(errorList)) + " folders were not able to be found.")
input("Press any key to exit...")
sys.exit()
| true |
77ee96c305f1d7d21ddae8c1029639a50627382b | SamuelHealion/cp1404practicals | /prac_05/practice_and_extension/electricity_bill.py | 1,317 | 4.1875 | 4 | """
CP1404 Practice Week 5
Calculate the electricity bill based on provided cents per kWh, daily use and number of billing days
Changed to use dictionaries for the tariffs
"""
TARIFFS = {11: 0.244618, 31: 0.136928, 45: 0.385294, 91: 0.374825, 33: 0.299485}
print("Electricity bill estimator 2.0")
print("Which tariff are you on?")
for tariff, cost in TARIFFS.items():
print("Tariff {} at {} per kWh".format(tariff, cost))
valid_tariff = False
while not valid_tariff:
try:
which_tariff = int(input(">>> "))
if which_tariff in TARIFFS.keys():
valid_tariff = True
else:
print("Not a valid tariff")
except ValueError:
print("Please enter a number")
daily_use = float(input("Enter daily use in kWh: "))
number_of_days = float(input("Enter number of billing days: "))
estimated_bill = TARIFFS[which_tariff] * daily_use * number_of_days
print("Estimated bill: ${:.2f}".format(estimated_bill))
"""
Original version
print("Electricity bill estimator")
cents_per_kwh = float(input("Enter cents per kWh: "))
daily_use = float(input("Enter daily use in kWh: "))
number_of_days = float(input("Enter number of billing days: "))
estimated_bill = (cents_per_kwh / 100) * daily_use * number_of_days
print("Estimated bill: ${:.2f}".format(estimated_bill))
"""
| true |
823282e7460b9d11b4d4127fa68a87352a5543ce | SamuelHealion/cp1404practicals | /prac_02/practice_and_extension/word_generator.py | 2,154 | 4.25 | 4 | """
CP1404/CP5632 - Practical
Random word generator - based on format of words
Another way to get just consonants would be to use string.ascii_lowercase
(all letters) and remove the vowels.
"""
import random
VOWELS = "aeiou"
CONSONANTS = "bcdfghjklmnpqrstvwxyz"
def first_version():
"""Requires c and v only"""
print("Please enter the word format: C for Consonants, V for Vowels")
word_format = str(input("> ").lower())
word = ""
for kind in word_format:
if kind == "c":
word += random.choice(CONSONANTS)
elif kind == "v":
word += random.choice(VOWELS)
else:
print("Invalid input")
break
print(word)
def second_version():
"""Uses wild cards but also accepts regular inputs
# for vowels, % for consonants and * for either"""
print("Random word generator that includes your characters but uses:")
print("Enter # for Vowels, % for Consonants or * for either")
word_format = str(input("> "))
word = ""
for kind in word_format:
if kind == "%":
word += random.choice(CONSONANTS)
elif kind == "#":
word += random.choice(VOWELS)
elif kind == "*":
word += random.choice(CONSONANTS + VOWELS)
else:
word += kind
print(word)
def third_version():
"""Automatically generates the word_format for a random length between 1 and 20"""
word_length = random.randint(0, 20)
word_format = ''
for i in range(word_length):
word_format += random.choice('%' + '#')
word = ""
for kind in word_format:
if kind == "%":
word += random.choice(CONSONANTS)
elif kind == "#":
word += random.choice(VOWELS)
print(word)
user_input = ''
while user_input != 'Q':
print("Press Q to quit")
print("What version would you like to run? 1, 2 or 3")
user_input = str(input(">>> ")).upper()
if user_input == '1':
first_version()
elif user_input == '2':
second_version()
elif user_input == '3':
third_version()
else:
print("Invalid option")
| true |
d2539727c20ffae59e81cccadb78648b10797a5d | SamuelHealion/cp1404practicals | /prac_06/guitar.py | 730 | 4.3125 | 4 | """
CP1404 Practical 6 - Classes
Define the class Guitar
"""
VINTAGE_AGE = 50
CURRENT_YEAR = 2021
class Guitar:
"""Represent a Guitar object."""
def __init__(self, name='', year=0, cost=0):
"""Initialise a Guitar instance."""
self.name = name
self.year = year
self.cost = cost
def __str__(self):
"""Return a string representation of a Guitar object."""
return "{} ({}) : ${:,.2f}".format(self.name, self.year, self.cost)
def get_age(self):
"""Return the age of the guitar."""
return CURRENT_YEAR - self.year
def is_vintage(self):
"""Return whether a Guitar is 50 years old or older."""
return self.get_age() >= VINTAGE_AGE
| true |
6339d40839889e191b3ef8dae558cf3266b08ba8 | jamieboyd/neurophoto2018 | /code/simple_loop.py | 407 | 4.46875 | 4 | #! /usr/bin/python
#-*-coding: utf-8 -*-
"""
a simple for loop with conditionals
% is the modulus operator, giving the remainder of the
integer division of the left operand by the right operand.
If a number divides by two with no remainder it is even.
"""
for i in range (0,10,1):
if i % 2 == 1:
print (str (i) + ' is an odd number.')
else:
print (str (i) + ' is an even number.')
| true |
86dec63990bd3ae55a023380af8ce0f38fcdf3e2 | CQcodes/MyFirstPythonProgram | /Main.py | 341 | 4.25 | 4 | import Fibonacci
import Check
# Driver Code
print("Program to print Fibonacci series upto 'n'th term.")
input = input("Enter value for 'n' : ")
if(Check.IsNumber(input)):
print("Printing fibonacci series upto '" + input + "' terms.")
Fibonacci.Print(int(input))
else:
print("Entered value is not a valid integer. Please Retry.")
| true |
f193bf10635f1b1cc9f4f7aa0ae7a209e5f041db | Yashs744/Python-Programming-Workshop | /if_else Ex-3.py | 328 | 4.34375 | 4 | # Nested if-else
name = input('What is your name? ')
# There we can provide any name that we want to check with
if name.endswith('Sharma'):
if name.startswith('Mr.'):
print ('Hello,', name)
elif name.startswith('Mrs.'):
print ('Hello,', name)
else:
print ('Hello,', name)
else:
print ('Hello, Stranger') | true |
012cc0af4adbfa714a4f311b729d89a9ba446d35 | Tagirijus/ledger-expenses | /general/date_helper.py | 1,213 | 4.1875 | 4 | import datetime
from dateutil.relativedelta import relativedelta
def calculateMonths(period_from, period_to):
"""Calculate the months from two given dates.."""
if period_from is False or period_to is False:
return 12
delta = relativedelta(period_to, period_from)
return abs((delta.years * 12) + delta.months)
def interpreteDate(date):
"""Interprete given date and return datetime or bool."""
if date is False:
return False
try:
return datetime.datetime.strptime(date, '%Y-%m')
except Exception as e:
return False
def normalizePeriods(months, period_from, period_to):
"""
Normlize the periods so that teh programm can work with them.
Basically it tries to generate the other period with self.months months
in difference, if only one period date is given.
"""
if period_from is False and period_to is False:
return (False, False)
elif period_from is False and period_to is not False:
period_from = period_to - relativedelta(months=months)
elif period_from is not False and period_to is False:
period_to = period_from + relativedelta(months=months)
return (period_from, period_to)
| true |
b5a66fd0978895aaabb5cb93de5a7cfabd57ad8e | yosef8234/test | /python_simple_ex/ex28.py | 477 | 4.3125 | 4 | # Write a function find_longest_word() that takes a list of words and
# returns the length of the longest one.
# Use only higher order functions.
def find_longest_word(words):
'''
words: a list of words
returns: the length of the longest one
'''
return max(list(map(len, words)))
# test
print(find_longest_word(["i", "am", "pythoning"]))
print(find_longest_word(["hello", "world"]))
print(find_longest_word(["ground", "control", "to", "major", "tom"]))
| true |
b787971db2b58732d63ea00aaac8ef233068b822 | yosef8234/test | /python_simple_ex/ex15.py | 443 | 4.25 | 4 | # Write a function find_longest_word() that takes a list of words and
# returns the length of the longest one.
def find_longest_word(words):
longest = ""
for word in words:
if len(word) >= len(longest):
longest = word
return longest
# test
print(find_longest_word(["i", "am", "pythoning"]))
print(find_longest_word(["hello", "world"]))
print(find_longest_word(["ground", "control", "to", "major", "tom"]))
| true |
9b22d6e5f777384cd3b88f9d44b7f2711346fc74 | yosef8234/test | /pfadsai/07-searching-and-sorting/notes/sequential-search.py | 1,522 | 4.375 | 4 | # Sequential Search
# Check out the video lecture for a full breakdown, in this Notebook all we do is implement Sequential Search for an Unordered List and an Ordered List.
def seq_search(arr,ele):
"""
General Sequential Search. Works on Unordered lists.
"""
# Start at position 0
pos = 0
# Target becomes true if ele is in the list
found = False
# go until end of list
while pos < len(arr) and not found:
# If match
if arr[pos] == ele:
found = True
# Else move one down
else:
pos = pos+1
return found
arr = [1,9,2,8,3,4,7,5,6]
print seq_search(arr,1) #True
print seq_search(arr,10) #False
# Ordered List
# If we know the list is ordered than, we only have to check until we have found the element or an element greater than it.
def ordered_seq_search(arr,ele):
"""
Sequential search for an Ordered list
"""
# Start at position 0
pos = 0
# Target becomes true if ele is in the list
found = False
# Stop marker
stopped = False
# go until end of list
while pos < len(arr) and not found and not stopped:
# If match
if arr[pos] == ele:
found = True
else:
# Check if element is greater
if arr[pos] > ele:
stopped = True
# Otherwise move on
else:
pos = pos+1
return found
arr.sort()
ordered_seq_search(arr,3) #True
ordered_seq_search(arr,1.5) #False | true |
720f5fa949f49b1fa20d7c0ae08ae397fc6fc225 | yosef8234/test | /pfadsai/03-stacks-queues-and-deques/notes/implementation-of-stack.py | 1,537 | 4.21875 | 4 | # Implementation of Stack
# Stack Attributes and Methods
# Before we implement our own Stack class, let's review the properties and methods of a Stack.
# The stack abstract data type is defined by the following structure and operations. A stack is structured, as described above, as an ordered collection of items where items are added to and removed from the end called the “top.” Stacks are ordered LIFO. The stack operations are given below.
# Stack() creates a new stack that is empty. It needs no parameters and returns an empty stack.
# push(item) adds a new item to the top of the stack. It needs the item and returns nothing.
# pop() removes the top item from the stack. It needs no parameters and returns the item. The stack is modified.
# peek() returns the top item from the stack but does not remove it. It needs no parameters. The stack is not modified.
# isEmpty() tests to see whether the stack is empty. It needs no parameters and returns a boolean value.
# size() returns the number of items on the stack. It needs no parameters and returns an integer.
class Stack:
def __init__(self):
self.items = []
def isEmpty(self):
return self.items == []
def push(self, item):
self.items.append(item)
def pop(self):
return self.items.pop()
def peek(self):
return self.items[len(self.items)-1]
def size(self):
return len(self.items)
s = Stack()
print(s.isEmpty())
s.push(1)
s.push('two')
s.peek()
s.push(True)
s.size()
s.isEmpty()
s.pop()
s.size() | true |
5d1b9a089f9f4c0e6b8674dafffa486f4698cdb4 | yosef8234/test | /toptal/python-interview-questions/4.py | 1,150 | 4.5 | 4 | # Q:
# What will be the output of the code below in Python 2? Explain your answer.
# Also, how would the answer differ in Python 3 (assuming, of course, that the above print statements were converted to Python 3 syntax)?
def div1(x,y):
print "%s/%s = %s" % (x, y, x/y)
def div2(x,y):
print "%s//%s = %s" % (x, y, x//y)
div1(5,2)
div1(5.,2)
div2(5,2)
div2(5.,2.)
# A:
# In Python 2, the output of the above code will be:
5/2 = 2
5.0/2 = 2.5
5//2 = 2
5.0//2.0 = 2.0
# By default, Python 2 automatically performs integer arithmetic if both operands are integers. As a result, 5/2 yields 2, while 5./2 yields 2.5.
# Note that you can override this behavior in Python 2 by adding the following import:
from __future__ import division
# Also note that the “double-slash” (//) operator will always perform integer division, regardless of the operand types. That’s why 5.0//2.0 yields 2.0 even in Python 2.
# Python 3, however, does not have this behavior; i.e., it does not perform integer arithmetic if both operands are integers. Therefore, in Python 3, the output will be as follows:
5/2 = 2.5
5.0/2 = 2.5
5//2 = 2
5.0//2.0 = 2.0 | true |
3c39888213a9dcc78c1c0a641b9ab70c87680c0a | yosef8234/test | /pfadsai/04-linked-lists/questions/linked-list-reversal.py | 2,228 | 4.46875 | 4 | # Problem
# Write a function to reverse a Linked List in place. The function will take in the head of the list as input and return the new head of the list.
# You are given the example Linked List Node class:
class Node(object):
def __init__(self,value):
self.value = value
self.nextnode = None
# Solution
# Since we want to do this in place we want to make the funciton operate in O(1) space, meaning we don't want to create a new list, so we will simply use the current nodes! Time wise, we can perform the reversal in O(n) time.
# We can reverse the list by changing the next pointer of each node. Each node's next pointer should point to the previous node.
# In one pass from head to tail of our input list, we will point each node's next pointer to the previous element.
# Make sure to copy current.next_node into next_node before setting current.next_node to previous. Let's see this solution coded out:
def reverse(head):
# Set up current,previous, and next nodes
current = head
previous = None
nextnode = None
# until we have gone through to the end of the list
while current:
# Make sure to copy the current nodes next node to a variable next_node
# Before overwriting as the previous node for reversal
nextnode = current.nextnode
# Reverse the pointer ot the next_node
current.nextnode = previous
# Go one forward in the list
previous = current
current = nextnode
return previous
# You should be able to easily test your own solution to make sure it works. Given the short list a,b,c,d with values 1,2,3,4. Check the effect of your reverse function and maek sure the results match the logic here below:
# Create a list of 4 nodes
a = Node(1)
b = Node(2)
c = Node(3)
d = Node(4)
# Set up order a,b,c,d with values 1,2,3,4
a.nextnode = b
b.nextnode = c
c.nextnode = d
print(a.nextnode.value)
print(b.nextnode.value)
print(c.nextnode.value)
reverse(a)
print(d.nextnode.value)
print(c.nextnode.value)
print(b.nextnode.value)
# Great, now we can see that each of the values points to its previous value (although now that the linked list is reversed we can see the ordering has also reversed)
| true |
9041e5cb16518965f170e82bdac4929e93ab0273 | yosef8234/test | /hackerrank/30-days-of-code/day-24.py | 2,826 | 4.375 | 4 | # # -*- coding: utf-8 -*-
# Objective
# Check out the Tutorial tab for learning materials and an instructional video!
# Task
# A Node class is provided for you in the editor. A Node object has an integer data field, datadata, and a Node instance pointer, nextnext, pointing to another node (i.e.: the next node in a list).
# A removeDuplicates function is declared in your editor, which takes a pointer to the headhead node of a linked list as a parameter. Complete removeDuplicates so that it deletes any duplicate nodes from the list and returns the head of the updated list.
# Note: The headhead pointer may be null, indicating that the list is empty. Be sure to reset your nextnext pointer when performing deletions to avoid breaking the list.
# Input Format
# You do not need to read any input from stdin. The following input is handled by the locked stub code and passed to the removeDuplicates function:
# The first line contains an integer, NN, the number of nodes to be inserted.
# The NN subsequent lines each contain an integer describing the datadata value of a node being inserted at the list's tail.
# Constraints
# The data elements of the linked list argument will always be in non-decreasing order.
# Output Format
# Your removeDuplicates function should return the head of the updated linked list. The locked stub code in your editor will print the returned list to stdout.
# Sample Input
# 6
# 1
# 2
# 2
# 3
# 3
# 4
# Sample Output
# 1 2 3 4
# Explanation
# N=6N=6, and our non-decreasing list is {1,2,2,3,3,4}{1,2,2,3,3,4}. The values 22 and 33 both occur twice in the list, so we remove the two duplicate nodes. We then return our updated (ascending) list, which is {1,2,3,4}{1,2,3,4}.
class Node:
def __init__(self,data):
self.data = data
self.next = None
class Solution:
def insert(self,head,data):
p = Node(data)
if head==None:
head=p
elif head.next==None:
head.next=p
else:
start=head
while(start.next!=None):
start=start.next
start.next=p
return head
def display(self,head):
current = head
while current:
print(current.data,end=' ')
current = current.next
def removeDuplicates(self,head):
node = head
while node:
if node.next:
if node.data == node.next.data:
node.next = node.next.next
else:
node = node.next
else:
node = node.next
return head
mylist= Solution()
T=int(input())
head=None
for i in range(T):
data=int(input())
head=mylist.insert(head,data)
head=mylist.removeDuplicates(head)
mylist.display(head) | true |
756868b809716f135bb1a27a9c35791e116a902a | yosef8234/test | /pfadsai/04-linked-lists/questions/implement-a-linked-list.py | 952 | 4.46875 | 4 | # Implement a Linked List - SOLUTION
# Problem Statement
# Implement a Linked List by using a Node class object. Show how you would implement a Singly Linked List and a Doubly Linked List!
# Solution
# Since this is asking the same thing as the implementation lectures, please refer to those video lectures and notes for a full explanation. The code from those lectures is displayed below:
class LinkedListNode(object):
def __init__(self,value):
self.value = value
self.nextnode = None
a = LinkedListNode(1)
b = LinkedListNode(2)
c = LinkedListNode(3)
a.nextnode = b
b.nextnode = c
class DoublyLinkedListNode(object):
def __init__(self,value):
self.value = value
self.next_node = None
self.prev_node = None
a = DoublyLinkedListNode(1)
b = DoublyLinkedListNode(2)
c = DoublyLinkedListNode(3)
# Setting b after a
b.prev_node = a
a.next_node = b
# Setting c after a
b.next_node = c
c.prev_node = b | true |
6eb333b658f76812126d0fefdb33a39e66f608bf | yosef8234/test | /pfadsai/10-mock-interviews/ride-share-company/on-site-question3.py | 2,427 | 4.3125 | 4 | # On-Site Question 3 - SOLUTION
# Problem
# Given a binary tree, check whether it’s a binary search tree or not.
# Requirements
# Use paper/pencil, do not code this in an IDE until you've done it manually
# Do not use built-in Python libraries to do this, but do mention them if you know about them
# Solution
# The first solution that comes to mind is, at every node check whether its value is larger than or equal to its left child and smaller than or equal to its right child (assuming equals can appear at either left or right). However, this approach is erroneous because it doesn’t check whether a node violates any condition with its grandparent or any of its ancestors.
# So, we should keep track of the minimum and maximum values a node can take. And at each node we will check whether its value is between the min and max values it’s allowed to take. The root can take any value between negative infinity and positive infinity. At any node, its left child should be smaller than or equal than its own value, and similarly the right child should be larger than or equal to. So during recursion, we send the current value as the new max to our left child and send the min as it is without changing. And to the right child, we send the current value as the new min and send the max without changing.
class Node:
def __init__(self, val=None):
self.left, self.right, self.val = None, None, val
INFINITY = float("infinity")
NEG_INFINITY = float("-infinity")
def isBST(tree, minVal=NEG_INFINITY, maxVal=INFINITY):
if tree is None:
return True
if not minVal <= tree.val <= maxVal:
return False
return isBST(tree.left, minVal, tree.val) and isBST(tree.right, tree.val, maxVal)
# There’s an equally good alternative solution. If a tree is a binary search tree, then traversing the tree inorder should lead to sorted order of the values in the tree. So, we can perform an inorder traversal and check whether the node values are sorted or not.
def isBST2(tree, lastNode=[NEG_INFINITY]):
if tree is None:
return True
if not isBST2(tree.left, lastNode):
return False
if tree.val < lastNode[0]:
return False
lastNode[0]=tree.val
return isBST2(tree.right, lastNode)
# This is a common interview problem, its relatively simple, but not trivial and shows that someone has a knowledge of binary search trees and tree traversals. | true |
0d349d0708bdb56ce5da09f585eaf41d8f9952c3 | yosef8234/test | /python_essential_q/q8.py | 2,234 | 4.6875 | 5 | # Question 8
What does this stuff mean: *args, **kwargs? And why would we use it?
# Answer
# Use *args when we aren't sure how many arguments are going to be passed to a function, or if we want to pass a stored list or tuple of arguments to a function. **kwargs is used when we dont know how many keyword arguments will be passed to a function, or it can be used to pass the values of a dictionary as keyword arguments. The identifiers args and kwargs are a convention, you could also use *bob and **billy but that would not be wise.
# Here is a little illustration:
def f(*args,**kwargs):
print(args, kwargs)
l = [1,2,3]
t = (4,5,6)
d = {'a':7,'b':8,'c':9}
f()
f(1,2,3) # (1, 2, 3) {}
f(1,2,3,"groovy") # (1, 2, 3, 'groovy') {}
f(a=1,b=2,c=3) # () {'a': 1, 'c': 3, 'b': 2}
f(a=1,b=2,c=3,zzz="hi") # () {'a': 1, 'c': 3, 'b': 2, 'zzz': 'hi'}
f(1,2,3,a=1,b=2,c=3) # (1, 2, 3) {'a': 1, 'c': 3, 'b': 2}
f(*l,**d) # (1, 2, 3) {'a': 7, 'c': 9, 'b': 8}
f(*t,**d) # (4, 5, 6) {'a': 7, 'c': 9, 'b': 8}
f(1,2,*t) # (1, 2, 4, 5, 6) {}
f(q="winning",**d) # () {'a': 7, 'q': 'winning', 'c': 9, 'b': 8}
f(1,2,*t,q="winning",**d) # (1, 2, 4, 5, 6) {'a': 7, 'q': 'winning', 'c': 9, 'b': 8}
def f2(arg1,arg2,*args,**kwargs):
print(arg1,arg2, args, kwargs)
f2(1,2,3) # 1 2 (3,) {}
f2(1,2,3,"groovy") # 1 2 (3, 'groovy') {}
f2(arg1=1,arg2=2,c=3) # 1 2 () {'c': 3}
f2(arg1=1,arg2=2,c=3,zzz="hi") # 1 2 () {'c': 3, 'zzz': 'hi'}
f2(1,2,3,a=1,b=2,c=3) # 1 2 (3,) {'a': 1, 'c': 3, 'b': 2}
f2(*l,**d) # 1 2 (3,) {'a': 7, 'c': 9, 'b': 8}
f2(*t,**d) # 4 5 (6,) {'a': 7, 'c': 9, 'b': 8}
f2(1,2,*t) # 1 2 (4, 5, 6) {}
f2(1,1,q="winning",**d) # 1 1 () {'a': 7, 'q': 'winning', 'c': 9, 'b': 8}
f2(1,2,*t,q="winning",**d) # 1 2 (4, 5, 6) {'a': 7, 'q': 'winning', 'c': 9, 'b': 8}
# Why do we care?
# Sometimes we will need to pass an unknown number of arguments or keyword arguments into a function. Sometimes we will want to store arguments or keyword arguments for later use. Sometimes it's just a time saver. | true |
ec8296cf056afef1f3ad97123c852b66f25d75cb | yosef8234/test | /pfadsai/04-linked-lists/notes/singly-linked-list-implementation.py | 1,136 | 4.46875 | 4 | # Singly Linked List Implementation
# In this lecture we will implement a basic Singly Linked List.
# Remember, in a singly linked list, we have an ordered list of items as individual Nodes that have pointers to other Nodes.
class Node(object):
def __init__(self,value):
self.value = value
self.nextnode = None
# Now we can build out Linked List with the collection of nodes:
a = Node(1)
b = Node(2)
c = Node(3)
a.nextnode = b
b.nextnode = c
# In a Linked List the first node is called the head and the last node is called the tail. Let's discuss the pros and cons of Linked Lists:
# Pros
# Linked Lists have constant-time insertions and deletions in any position, in comparison, arrays require O(n) time to do the same thing.
# Linked lists can continue to expand without having to specify their size ahead of time (remember our lectures on Array sizing form the Array Sequence section of the course!)
# Cons
# To access an element in a linked list, you need to take O(k) time to go from the head of the list to the kth element. In contrast, arrays have constant time operations to access elements in an array. | true |
09da1340b227103c7eb1bc9800c714907939bfde | yosef8234/test | /python_ctci/q1.4_permutation_of_palindrom.py | 1,097 | 4.21875 | 4 | # Write a function to check if a string is a permutation of a palindrome.
# Permutation it is "abc" == "cba"
# Palindrome it is "Madam, I'm Adam'
# A palindrome is word or phrase that is the same backwards as it is forwards. (Not limited to dictionary words)
# A permutation is a rearrangement of letters.
import string
def isPermutationOfPalindrome(str):
# {'a': False, 'c': False, 'b': False, 'e': False, 'd': False, 'g': False, 'f': False, 'i': False, 'h': False, 'k': False, 'j': False, 'm': False, 'l': False, 'o': False, 'n': False, 'q': False, 'p': False, 's': False, 'r': False, 'u': False, 't': False, 'w': False, 'v': False, 'y': False, 'x': False, 'z': False}
d = dict.fromkeys(string.ascii_lowercase, False)
count = 0
for char in str:
if(ord(char) > 96 and ord(char) < 123):
d[char] = not d[char]
for key in d:
if d[key] is True:
count += 1
if count > 1:
return False
return True
print(isPermutationOfPalindrome("abcecba")) # True
print(isPermutationOfPalindrome("aa bb cc eee f")) # False | true |
c117f5f321a25493ee9c3811a51e6c28d6487392 | imjching/playground | /python/practice_python/11_check_primality_functions.py | 730 | 4.21875 | 4 | # http://www.practicepython.org/exercise/2014/04/16/11-check-primality-functions.html
"""
Ask the user for a number and determine whether the number
is prime or not. (For those who have forgotten, a prime number
is a number that has no divisors.). You can (and should!)
use your answer to
[Exercise 4](/exercise/2014/02/26/04-divisors.html to help you. Take this opportunity to practice using functions, described below.
"""
number = int(raw_input("Please enter a number: "))
def is_prime(number):
if number < 2:
return False
for i in range(2, (number / 2) + 1):
if number % i == 0:
return False
return True
if is_prime(number):
print "This is a prime number!"
else:
print "This is not a prime number"
| true |
8f13963a5059a9cbb14790645f86ff0415398108 | imjching/playground | /python/practice_python/14_list_remove_duplicates.py | 764 | 4.1875 | 4 | # http://www.practicepython.org/exercise/2014/05/15/14-list-remove-duplicates.html
"""
Write a program (function!) that takes a list and returns a new
list that contains all the elements of the first list minus all
the duplicates.
Extras:
Write two different functions to do this - one using a loop and
constructing a list, and another using sets.
Go back and do Exercise 5 using sets, and write the solution
for that in a different function.
"""
from sets import Set
def unique_elements_loop(_list):
new_list = []
for i in _list:
if i not in new_list:
new_list.append(i)
return new_list
def unique_elements_sets(_list):
return list(Set(_list))
print unique_elements_loop([1, 2, 3, 3, 3, 3])
print unique_elements_sets([1, 2, 3, 3, 3, 3])
| true |
35c0ab9c2e6bfb4eea6f3750b208495ce1407d03 | imjching/playground | /python/practice_python/18_cows_and_bulls.py | 1,813 | 4.21875 | 4 | # http://www.practicepython.org/exercise/2014/07/05/18-cows-and-bulls.html
"""
Create a program that will play the 'cows and bulls' game with the user.
The game works like this:
Randomly generate a 4-digit number. Ask the user to guess a 4-digit number.
For every digit that the user guessed correctly in the correct place,
they have a 'cow'.
For every digit the user guessed correctly in the wrong place is a 'bull'.
Every time the user makes a guess, tell them how many 'cows' and 'bulls'
they have. Once the user guesses the correct number, the game is over.
Keep track of the number of guesses the user makes throughout the game and
tell the user at the end.
Say the number generated by the computer is 1038.
An example interaction could look like this:
Welcome to the Cows and Bulls Game!
Enter a number:
>>> 1234
2 cows, 0 bulls
>>> 1256
1 cow, 1 bull
...
Until the user guesses the number.
"""
from random import randint
print "Welcome to the Cows and Bulls Game!"
number = "".join(["1234567890"[randint(0, 9)] for i in range(4)])
print number
def check(input_, number):
cows = 0
bulls = 0
for i in range(len(number)):
if input_[i] == number[i]:
cows += 1
elif input_[i] in number: # something doesn't seem right
bulls += 1
if cows == 4:
return "0"
else:
res = str(cows) + " cow, " if cows <= 1 else str(cows) + " cows, "
res += str(bulls) + " bull" if bulls <= 1 else str(bulls) + " bulls"
return res
trials = 0
while True:
trials += 1
input_ = raw_input("Enter a number: ")
if len(input_) != 4:
print "Please enter a 4 digit number."
else:
result = check(input_, number)
if result == "0":
print "You have guessed the number with a total of " + str(trials) + " guesses!"
break
else:
print result
| true |
53937a32b059e4e9613be47b492f586eff09a06d | bkhuong/LeetCode-Python | /make_itinerary.py | 810 | 4.125 | 4 | class Solution:
'''
Given a list of tickets, find itinerary in order using the given list.
'''
def find_route(tickets:list) -> str:
routes = {}
start = []
# create map
for ticket in tickets:
routes[ticket[0]] = {'to':ticket[1]}
try:
routes[ticket[1]].update({'from':ticket[0]})
except:
routes[ticket[1]] = {'from':ticket[0]}
# find starting point
for k,v in routes.items():
if 'from' not in v.keys():
departure = k
break
# walk dictionary for route:
while 'to' in routes[departure].keys():
print(f'{departure} -> {routes[departure]["to"]}')
departure = routes[departure]['to']
| true |
70cc3581b224daa3beadfc0150d31b52c30f6284 | Gachiman/Python-Course | /python-scripts/hackerrank/Medium/Find Angle MBC.py | 603 | 4.21875 | 4 | import math
def input_length_side(val):
while True:
try:
length = int(float(input("Enter the length of side {0} (0 < {0} <= 100): ".format(val))))
if 0 < length <= 100:
return length
else:
raise ValueError
except ValueError:
print("Please reinsert")
def main():
side_ab = input_length_side("AB")
side_bc = input_length_side("BC")
tg_mbc = side_ab / side_bc
angle_mbc = math.degrees(math.atan(tg_mbc))
print(round(angle_mbc), '°', sep='')
if __name__ == '__main__':
main()
| true |
a1b6391a773b23a0f5fe8e0b0a4d36bc7e03b9b0 | hobsond/Computer-Architecture | /white.py | 1,788 | 4.625 | 5 | # Given the following array of values, print out all the elements in reverse order, with each element on a new line.
# For example, given the list
# [10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
# Your output should be
# 0
# 1
# 2
# 3
# 4
# 5
# 6
# 7
# 8
# 9
# 10
# You may use whatever programming language you'd like.
# Verbalize your thought process as much as possible before writing any code. Run through the UPER problem solving framework while going through your thought process.
originalList = [10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
reverseList = originalList[::-1]
for i in reverseList:
print(i)
# Given a hashmap where the keys are integers, print out all of the values of the hashmap in reverse order, ordered by the keys.
# For example, given the following hashmap:
# {
# 14: "vs code",
# 3: "window",
# 9: "alloc",
# 26: "views",
# 4: "bottle",
# 15: "inbox",
# 79: "widescreen",
# 16: "coffee",
# 19: "tissue",
# }
# The expected output is:
# widescreen
# views
# tissue
# coffee
# inbox
# vs code
# alloc
# bottle
# window
# since "widescreen" has the largest integer key, "views" has the second largest, etc.
# You may use whatever programming language you'd like.
# Verbalize your thought process as much as possible before writing any code. Run through the UPER problem solving framework while going through your thought process.
# hash map
key = {
14: "vs code",
3: "window",
9: "alloc",
26: "views",
4: "bottle",
15: "inbox",
79: "widescreen",
16: "coffee",
19: "tissue",
}
# loop through key get the key.items()
# which im gonna sort by the first value
# then loop through and print
newList = sorted([i for i in key.items()],reverse=True)
# newList.sort(key = lambda e:e[0])
# newList.reverse()
for i in newList:
print(i[1]) | true |
2718e25886a29226c5050afe0d83c6459bd6747b | Twest19/prg105 | /Ch 10 HW/10-1_person_data.py | 1,659 | 4.71875 | 5 | """
Design a class that holds the following personal data: name, address, age, and phone number.
Write appropriate accessor and mutator methods (get and set). Write a program that creates three instances
of the class. One instance should hold your information and the other two should hold your friends' or
family members' information. Just add information, don't get it from the user. Print the data from each
object, make sure to format the output for clarity and ease of reading.
"""
class Data: # This class takes the data for a person and returns that data
def __init__(self, name, address, age, phone='n/a'):
self.name = name
self.address = address
self.age = age
self.phone = phone
def set_name(self, name):
self.name = name
def get_name(self):
return self.name
def set_address(self, address):
self.address = address
def get_address(self):
return self.address
def set_age(self, age):
self.age = int(age)
def get_age(self):
return self.age
def set_phone(self, phone):
self.phone = phone
def get_phone(self):
return self.phone
def __str__(self):
return f"Name: {self.name}\nAddress: {self.address}\nAge: {self.age}\nPhone: {self.phone}\n"
def main():
my_data = Data('Tim', '119 W South Ave', 21, '815-354-7216')
friend_data = Data('Saul', '9800 Montgomery Blvd', 49, '505-503-4455')
family_data = Data('Tom', '191 E North Ave', 42)
print(my_data)
print(friend_data)
print(family_data)
main()
| true |
b33b65d4b831bcd41fac9f4cd424a65dc4589d39 | Twest19/prg105 | /3-3_ticket.py | 2,964 | 4.3125 | 4 | """
You are writing a program to sell tickets to the school play.
If the person buying the tickets is a student, their price is $5.00 per ticket.
If the person buying the tickets is a veteran, their price is $7.00 per ticket.
If the person buying the ticket is a sponsor of the play, the price is $2.00 per ticket.
"""
# Must initialize price and discount to later assign them a new value
price = 0
discount = 0
print(f"{'=' * 6} Play Ticket Types w/ Prices {'=' * 6}"
"\n(1) Student - $6.00"
"\n(2) Veteran - $7.00"
"\n(3) Show Sponsor - $2.00"
"\n(4) Retiree - $6.00"
"\n(5) General Public - $10.00")
print(f"{'=' * 36}")
# Ask the buyer which ticket group they fall into
# Choose a number 1-5 or it will display an error and prompt them until they provide a proper response
while True:
prompt = input("\nPlease choose a number for the corresponding ticket type that you fall into: ")
if prompt in ('1', '2', '3', '4', '5'):
ticket_type = int(prompt)
# Assign the ticket to a price for the number chosen by the buyer
if ticket_type == 1:
price = 6.00
print("\nYou have selected Student.")
elif ticket_type == 2:
price = 7.00
print("\nYou have selected Veteran.")
elif ticket_type == 3:
price = 2.00
print("\nYou have selected Show Sponsor.")
elif ticket_type == 4:
price = 6.00
print("\nYou have selected Retiree.")
elif ticket_type == 5:
price = 10.00
print("\nYou have selected General Public.")
break
else:
print("\nError please try again")
# Let the buyer know that they can receive a discount when they buy more tickets
print("\nIf you buy 4 - 8 tickets you get a 10% discount.")
print("If you buy 9 - 15 tickets you get a 15% discount.")
print("If you buy more than 15 tickets you get a 20% discount.")
# Ask buyer how many tickets they would like to purchase
ticket_quantity = int(input("\nHow many tickets would you like to buy? "))
# Get the total price before any discounts are applied if there are any
ticket_total = price * ticket_quantity
# Determine if the buyer has met the requirements for a discount
if ticket_quantity > 15:
print("\nYou get a 20% discount!")
discount = ticket_total * .20
elif ticket_quantity >= 9:
print("\nYou get a 15% discount!")
discount = ticket_total * .15
elif ticket_quantity >= 4:
print("\nYou get a 10% discount")
discount = ticket_total * .10
else:
print("\nSorry, you do not get a discount.")
# Calculate the price by applying the discount
discount_applied = ticket_total - discount
# Calculate average price paid
avg = discount_applied / ticket_quantity
# Display the total before and after discount with how much they saved
print(f"\nTotal before discount: ${ticket_total:.2f}")
print(f"You saved: ${discount:.2f} ")
print(f"Your price per ticket is: ${avg:.2f}")
print(f"Total after discount: ${discount_applied:.2f}")
| true |
20a0d53d34ba73884e14d026030289475bb6275e | Twest19/prg105 | /chapter_practice/ch_9_exercises.py | 2,681 | 4.4375 | 4 | """
Complete all of the TODO directions
The number next to the TODO represents the chapter
and section in your textbook that explain the required code
Your file should compile error free
Submit your completed file
"""
import pickle
# TODO 9.1 Dictionaries
print("=" * 10, "Section 9.1 dictionaries", "=" * 10)
# 1) Finish creating the following dictionary by adding three more people and birthdays
birthdays = {'Meri': 'May 16', 'Kathy': 'July 14'}
birthdays['Tim'] = 'Jan 19'
birthdays['Jane'] = 'Jun 30'
birthdays['Harold'] = 'Oct 19'
# 2) Print Meri's Birthday
print(birthdays['Meri'])
# 3) Create an empty dictionary named registration
registration = {}
# You will use the following dictionary for many of the remaining exercises"
miles_ridden = {'June 1': 25, 'June 2': 20, 'June 3': 38, 'June 4': 12, 'June 5': 30, 'June 7': 25}
# 1) Print the keys and the values of miles_ridden using a for loop
for k, v in miles_ridden.items():
print(k, v)
# 2) Get the value for June 3 and print it, if not found display 'Entry not found', replace the ""
value = miles_ridden['June 3']
print(value)
# 3) Get the value for June 6 and print it, if not found display 'Entry not found' replace the ""
if 'June 6' in miles_ridden:
value2 = miles_ridden['June 6']
else:
value2 = 'Entry not found'
print(value2)
# 4) Use the values method to print the miles_ridden dictionary
print(miles_ridden.values())
# 5) Use the keys method to print all of the keys in miles_ridden
print(miles_ridden.keys())
# 6) Use the pop method to remove June 4 then print the contents of the dictionary
miles_ridden.pop('June 4')
print(miles_ridden)
# 7) Use the items method to print the contents of the miles_ridden dictionary
print(miles_ridden.items())
# TODO 9.2 Sets
print("=" * 10, "Section 9.2 sets", "=" * 10)
# 1) Create an empty set named my_set
my_set = set()
# 2) Create a set named days that contains the days of the week
days = {'Sun', 'Mon', 'Tues', 'Wed', 'Thurs', 'Fri', 'Sat'}
# 3) Get the number of elements from the days set and print it
print(len(days))
# 4) Remove Saturday and Sunday from the days set
days.remove('Sun')
days.remove('Sat')
print(days)
# 5) Determine if 'Mon' is in the days set
if 'Mon' in days:
print('Mon')
# TODO 9.3 Serializing Objects (Pickling)
print("=" * 10, "Section 9.3 serializing objects using the pickle library", "=" * 10)
# 1) Import the pickle library at the top of this file
# 2) Create the output file log and open it for binary writing
output = open('log.dat', 'wb')
# 3) Pickle the miles_ridden dictionary and output it to the log file
pickle.dump(miles_ridden, output)
# 4) Close the log file
output.close()
| true |
666efab8a625d46543dab413aadd15936594a5dd | Twest19/prg105 | /4-1_sales.py | 862 | 4.5625 | 5 | """
You need to create a program that will have the user enter in the total sales amount for the day at a coffee shop.
The program should ask the user for the total amount of sales and include the day in the request. At the end of
data entry, tell the user the total sales for the week, and the average sales per day. You must create a list of
the days of the week for the user to step through, see the example output.
"""
days = ['Sunday', 'Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday']
weekly_total = 0
print("Sales for the week:")
for day in days:
sales = float(input(f"What was the total amount of sales on {day}? "))
weekly_total += sales
print(f"\nThe total amount of sales for the week was: ${weekly_total:,.2f}")
print(F"The average amount of sales per day was: ${weekly_total / 7:,.2f}")
| true |
a3ace412c840aac7ff86999020c0d765a5062a5d | Twest19/prg105 | /5-3_assessment.py | 2,467 | 4.21875 | 4 | """
You are going to write a program that finds the area of a shape for the user.
"""
# set PI as a constant to be used as a global value
PI = 3.14
# create a main function then from the main function call other functions to get the correct calculations
def main():
while True:
menu()
choice = int(input("Enter the number of your selection: "))
num = validate(choice)
if num == 1:
base = int(input("Enter the base of the rectangle in cm: "))
height = int(input("Enter the height of the rectangle in cm: "))
area = rectangle(base, height)
print(f"The area of the rectangle is {area:.2f} square cm.")
elif num == 2:
base = int(input("Enter the base of the triangle in cm: "))
height = int(input("Enter the height of the triangle in cm: "))
area = triangle(base, height)
print(f"The area of the triangle is {area:.2f} square cm.")
elif num == 3:
side = int(input("Enter the length of one side of the square in cm: "))
area = square(side)
print(f"The area of the square is {area:.2f} square cm.")
elif num == 4:
radius = int(input("Enter the radius of the circle: "))
area = circle(radius)
print(f"The area of the circle is {area:.2f} square cm.")
# menu function displays the options available to the user
def menu():
print("\nThis program will find the area of any of the shapes below.")
print("1. Rectangle\n"
"2. Triangle\n"
"3. Square\n"
"4. Circle\n"
"5. Quit\n")
# The validate function confirms the users input if the users provides a correct response
def validate(x):
while True:
if 1 <= x < 5:
return x
elif x == 5:
print("\nGoodbye!")
exit()
else:
print("\nError, please try again.\n")
x = int(input("Enter the number of your selection: "))
continue
# Create a function for each of the shapes to calculate the area for the corresponding shape
def rectangle(x, y):
area = x * y
return area
def triangle(x, y):
area = x * y * 1/2
return area
def square(x):
area = x**2
return area
def circle(r):
global PI
area = PI * r**2
return area
main()
| true |
52dd577610c57f96e36b41ee06982c873f0d55af | dougiejim/Automate-the-boring-stuff | /commaCode.py | 753 | 4.3125 | 4 | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
#Write a function that takes a list value as an argument and returns
#a string with all the items separated by a comma and a space, with
#and inserted before the last item. For example, passing the previous
#spam list to the function would return 'apples, bananas, tofu, and cats'.
#But your function should be able to work with any list value passed to it.
"""
Created on Sat Jan 20 15:53:15 2018
@author: coledoug
"""
spam = ['apples', 'bananas', 'tofu', 'cats']
newString = ''
def stringReturn(list):
newString = ''
for i in list:
if i != list[-1]:
return newString + str(i +', ')
else:
return newString + str(' and '+ i)
stringReturn(spam)
| true |
732a68dc8a28c98ecc93001de996919759778a2c | sakamoto-michael/Sample-Python | /new/intro_loops.py | 727 | 4.28125 | 4 | # Python Loops and Iterations
nums = [1, 2, 3, 4, 5]
# Looping through each value in a list
for num in nums:
print(num)
# Finding a value in a list, breaking upon condition
for num in nums:
if num == 3:
print('Found!')
break
print(num)
# Finding a value, the continuing execution
for num in nums:
if num == 3:
print('Found')
continue
print(num)
# Loops within loops
for num in nums:
for letter in 'abc':
print(num, letter)
# Iterations - range(starting point, end point)
for i in range(1, 11):
print(i)
# While loop - must include a breaking condition. Can also included manual break within loop upon a different condition
x = 0
while x < 10:
if x == 5:
break
print(x)
x += 1
| true |
701ea9976f04c66564962c3bc7f64d89e1314120 | vivekyadav6838/Data-Structures-and-Algorithms-for-Interviews | /Python/BinaryTree/NumberOfNonLeafNodes.py | 1,524 | 4.3125 | 4 | # @author
# Aakash Verma
# Output:
# Pre Order Traversal is: 1 2 4 5 3 6 7
# Number Of non-Leaf Nodes: 3
# Creating a structure for the node.
# Initializing the node's data upon calling its constructor.
class Node:
def __init__(self, data):
self.data = data
self.right = self.left = None
# Defining class for the Binary Tree
class NumberOfNonLeafNodes:
# Assigning root as null initially. So as soon as the object will be created
# of this NumberOfNonLeafNodes class, root will be set as null.
def __init__(self):
self.root = None
# Pre Order Traversal
def preOrder(self, root):
if root is None:
return
print(root.data, end = " ")
self.preOrder(root.left)
self.preOrder(root.right)
def numNonLeafNodes(self, root):
if root is None:
return 0
if root.left is None and root.right is None:
return 0;
return 1 + self.numNonLeafNodes(root.left) + self.numNonLeafNodes(root.right)
# main method
if __name__ == '__main__':
tree = NumberOfNonLeafNodes()
tree.root = Node(1)
tree.root.left = Node(2)
tree.root.right = Node(3)
tree.root.left.left = Node(4)
tree.root.left.right = Node(5)
tree.root.right.left = Node(6)
tree.root.right.right = Node(7)
print("Pre Order Traversal is:", end = " ")
tree.preOrder(tree.root)
print()
print("Number Of non-Leaf Nodes:", end = " ")
print(tree.numNonLeafNodes(tree.root))
| true |
8878c006639c546777ff2af254a979033560c15a | vivekyadav6838/Data-Structures-and-Algorithms-for-Interviews | /Python/LinkedList/StartingOfLoop.py | 1,519 | 4.34375 | 4 | #
#
#
# @author
# Aakash Verma
#
# Start of a loop in Linked List
#
# Output:
# 5
#
# Below is the structute of a node which is used to create a new node every time.
class Node:
def __init__(self, data):
self.data = data
self.next = None # None is nothing but null
# Creating a class for implementing the code for Starting Of a loop in a LinkedList
class LinkedList:
# Whenever I'll create the object of a LinkedList class head will be pointing to null initially
def __init__(self):
self.head = None
# Inserting at the end of a Linked List (append function)
def append(self, key):
h = self.head
if h is None:
new_node = Node(key)
self.head = new_node
else:
while h.next != None:
h = h.next
new_node = Node(key)
h.next = new_node
# middle of a linked list
def startingOfLoop(self):
fast = self.head
slow = self.head
isLoopFound = False
if self.head is None:
print("The list doesn't exist.")
return
while fast is not None and fast.next is not None:
slow = slow.next
fast = fast.next.next
if slow == fast:
isLoopFound = True
break
if isLoopFound is True:
slow = self.head
while slow != fast:
slow = slow.next
fast = fast.next
print(slow.data)
# Code execution starts here
if __name__=='__main__':
myList = LinkedList()
myList.append(3)
myList.append(4)
myList.append(5)
myList.append(6)
myList.append(7)
myList.head.next.next.next.next.next = myList.head.next.next
myList.startingOfLoop()
| true |
96ca87b2c6191ffd17b5571581f5dec816529ef2 | SgtHouston/python102 | /sum the numbers.py | 249 | 4.21875 | 4 | # Make a list of numbers to sum
numbers = [1, 2, 3, 4, 5, 6]
# set up empty total so we can add to it
total = 0
# add current number to total for each number iun the list
for number in numbers:
total += number
# print the total
print(total)
| true |
25ceca21258ebec39c3bb51f308a1e5f136aaca4 | urmajesty/learning-python | /ex5.py | 581 | 4.15625 | 4 | name = 'Zed A Shaw'
age = 35.0 # not a lie
height = 74.0 # inches
weight = 180.0 #lbs
eyes = 'Blue'
teeth = 'White'
hair = 'Brown'
print(f"Let's talk about {name}.")
print(f"He's {height} pounds heavy.")
print("Actually that's not too heavy.")
print(f"He's got {eyes} eyes and {hair} hair.")
print(f"His teeth are usually {teeth} depending on the coffee.")
total = age + height + weight
print(f"If I add {age}, {height}, and {weight} I get {total}.")
#study drill #2
cm = (height * 2.54)
print(f"He also is {cm} cm tall.")
kg = (weight * .453592)
print(f"And He weighs {kg} kg.") | true |
53c52d5fb29d5f8f28c56ead8f8c31dfd6f06d98 | antoinemadec/test | /python/codewars/simplifying/simplifying.py | 2,602 | 4.5 | 4 | #!/usr/bin/env python3
'''
You are given a list/array of example formulas such as:
[ "a + a = b", "b - d = c ", "a + b = d" ]
Use this information to solve a formula in terms of the remaining symbol such as:
"c + a + b" = ?
in this example:
"c + a + b" = "2a"
Notes:
Variables names are case sensitive
There might be whitespaces between the different characters. Or not...
There should be support for parenthesis and their coefficient. Example: a +
3 (6b - c).
You might encounter several imbricated levels of parenthesis but you'll
never get a variable as coefficient for parenthesis, only constant terms.
All equations will be linear
See the sample tests for clarification of what exactly the input/ouput formatting should be.
Without giving away too many hints, the idea is to substitute the examples into
the formula and reduce the resulting equation to one unique term. Look
carefully at the example tests: you'll have to identify the pattern used to
replace variables in the formula/other equations. Only one solution is possible
for each test, using this pattern, so if you keep asking yourself "but what if
I do that instead of...", that is you missed the thing.
'''
import re
def simplify(examples,formula):
d,letters = {},[]
examples = [re.sub('([0-9]+) *([(a-zA-Z])',r'\1*\2',e) for e in examples]
for e in examples:
m = re.match('(?P<expr>.*) += +(?P<var>\w+)',e)
d[m.group('var')] = m.group('expr')
letters.extend([c for c in e if c.isalpha()])
c = [c for c in letters if c not in d][0]
d[c] = '1'
for _ in range(1000):
formula = re.sub('([0-9]+) *([(a-zA-Z])',r'\1*\2',formula)
for k in d:
formula = formula.replace(k,'(' + d[k] + ')')
try:
r = eval(formula)
except:
continue
return "%d%c" % (r,c)
examples=[["a + a = b", "b - d = c", "a + b = d"],
["a + 3g = k", "-70a = g"],
["-j -j -j + j = b"]]
formula=["c + a + b",
"-k + a",
"-j - b"
]
answer=["2a",
"210a",
"1j"
]
for i in range(len(answer)):
print('examples:' + str(examples[i]))
print('formula:' + str(formula[i]))
print('expected answer:'+str(answer[i]))
print(simplify(examples[i],formula[i]))
examples=['y + 6Y - k - 6 K = f', ' F + k + Y - y = K', 'Y = k', 'y = Y', 'y + Y = F']
formula='k - f + y'
print('expected answer:14y')
print(simplify(examples,formula))
examples=['x = b', 'b = c', 'c = d', 'd = e']
formula='c'
print('expected answer:1x')
print(simplify(examples,formula))
| true |
1d5eba8fd2834bb2375016ec7ed9e8cc686f1991 | antoinemadec/test | /python/programming_exercises/q5/q5.py | 662 | 4.21875 | 4 | #!/usr/bin/env python3
print("""https://raw.githubusercontent.com/zhiwehu/Python-programming-exercises/master/100%2B%20Python%20challenging%20programming%20exercises.txt
Question:
Define a class which has at least two methods:
getString: to get a string from console input
printString: to print the string in upper case.
Also please include simple test function to test the class methods.
Answer:""")
class String:
def __init__(self):
self.string = ""
def getString(self):
self.string = input("Enter string: ")
def printString(self):
print(self.string.upper())
s = String();
s.getString()
s.printString()
| true |
7f17dd1d0b9acc663a17846d838cbd79998bb79b | antoinemadec/test | /python/programming_exercises/q6/q6.py | 840 | 4.21875 | 4 | #!/usr/bin/env python3
print("""https://raw.githubusercontent.com/zhiwehu/Python-programming-exercises/master/100%2B%20Python%20challenging%20programming%20exercises.txt
Question:
Write a program that calculates and prints the value according to the given formula:
Q = Square root of [(2 * C * D)/H]
Following are the fixed values of C and H:
C is 50. H is 30.
D is the variable whose values should be input to your program in a comma-separated sequence.
Example
Let us assume the following comma separated input sequence is given to the program:
100,150,180
The output of the program should be:
18,22,24
Answer:""")
C = 50
H = 30
Q = []
for D in input("Enter numbers separated by comma: ").split(','):
D = int(D)
Q.append(int(((2*C*D)/H) ** 0.5))
print(','.join([str(i) for i in Q]))
| true |
d057707ccd873e895d2caf5eec45a19e0473da84 | antoinemadec/test | /python/codewars/find_the_divisors/find_the_divisors.py | 708 | 4.3125 | 4 | #!/usr/bin/env python3
"""
Create a function named divisors/Divisors that takes an integer and returns an
array with all of the integer's divisors(except for 1 and the number itself).
If the number is prime return the string '(integer) is prime' (null in C#) (use
Either String a in Haskell and Result<Vec<u32>, String> in Rust).
Example:
divisors(12); #should return [2,3,4,6]
divisors(25); #should return [5]
divisors(13); #should return "13 is prime"
You can assume that you will only get positive integers as inputs.
"""
def divisors(i):
r = [x for x in range(2,i) if (i//x)*x == i]
if r:
return r
else:
return "%0d is prime" % i
print(divisors(12))
print(divisors(13))
| true |
81cc9b7d03933d990e1a90129929d1eb78ffb108 | antoinemadec/test | /python/cs_dojo/find_mult_in_list/find_mult_in_list.py | 526 | 4.1875 | 4 | #!/bin/env python3
def find_multiple(int_list, multiple):
sorted_list = sorted(int_list)
n = len(sorted_list)
for i in range(0,n):
for j in range(i+1,n):
x = sorted_list[i]
y = sorted_list[j]
if x*y == multiple:
return (x,y)
elif x>multiple:
return None
return None
# execution
l = [int(x) for x in input("Enter list (white space separated): ").split(" ")]
m = int(input("Enter multiple: "))
print(find_multiple(l,m))
| true |
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