code/scripts/12_new_pairs.py

"""Collect candidates the rescue channels found that model-4 never scored, and shard them.

THE BUDGET IS THE WHOLE PROBLEM. Union the rescue channels naively and you get well over 100M
pairs. At roughly 2,400 pairs/s per GPU, four lanes clear about 35M pairs an hour, so an
unbounded union is a two-day job. Pairs are therefore selected, not merged:

  1. Drop everything model-4 already scored. Its test table holds 9,980,755 pairs with a
     cross-encoder score; re-scoring them buys nothing.
  2. Keep every hit from the rescue channels (F4/F5). These are the targeted ones - the whole
     point of building them - and they are few per S1 because k is small.
  3. From the wide e5-small k=100 channel, keep only ranks beyond what model-4 used. Its
     never-retrieved audit put 926 pairs (0.095 points) in fused rank 64-99, so the value
     there is real but bounded; spending the entire budget on it would be a bad trade
     against F4/F5, which target 0.42 points.
  4. Cap per S1. Without a cap a handful of entities in dense name clusters absorb the budget.

Everything is reported, so if the total is still too large for the time available you can
lower --per-s1 and re-run in seconds rather than discovering it three hours into scoring.
"""
from __future__ import annotations

import argparse
import glob
import os
import sys

import numpy as np
import pandas as pd

sys.path.insert(0, os.path.dirname(os.path.dirname(os.path.abspath(__file__))))
from berx import ce_score as CE  # noqa: E402
from berx import fuse as F  # noqa: E402
from berx import paths as P  # noqa: E402

# "ename" is the measured character-TF-IDF channel (44% recovery at k=5). "nameonly" is my
# earlier dense version of the same idea, kept because it is already computed and costs
# nothing to union in. "addronly" is deliberately absent: measured at 6-12% recovery for
# 10-50 candidates per S1 and rejected.
RESCUE = ("ename", "nameonly")


def load_channel(split: str, ch: str):
    fs = sorted(glob.glob(P.work("retr", f"{split}_{ch}_s*.parquet")))
    if not fs:
        return None
    d = pd.concat([pd.read_parquet(f) for f in fs], ignore_index=True)
    print(f"  {ch:<10} {len(d):>12,} pairs from {len(fs)} file(s)")
    return d


def main():
    ap = argparse.ArgumentParser()
    ap.add_argument("--split", default="test")
    ap.add_argument("--nshards", type=int, default=4)
    ap.add_argument("--wide-rank-min", type=int, default=64,
                    help="from the wide channel keep only ranks >= this")
    ap.add_argument("--per-s1", type=int, default=12, help="cap on NEW pairs per S1")
    ap.add_argument("--rescue-rank-max", type=int, default=5,
                    help="unmeasured rescue channels are capped to this many per S1, "
                         "matching the k=5 that was measured for ename")
    ap.add_argument("--max-total", type=int, default=60_000_000)
    a = ap.parse_args()

    have = pd.read_parquet(P.work("m4", f"{a.split}.parquet"), columns=["q", "c"])
    print(f"model-4 already scored {len(have):,} pairs")
    seen = set(map(tuple, have.to_numpy().tolist()))

    # PRIORITY BY STRENGTH OF EVIDENCE, not by how many pairs a channel happens to emit.
    #
    # The first run of this script gave ename and nameonly the same priority and a cap of 24.
    # Result: nameonly took 33.3M of the 41.6M budget and ename got 4.2M - so 80% of the GPU
    # hours would have gone to the channel with NO measured recovery rate, and 10% to the one
    # measured at 44%. A cap applied to a mixed pool silently ranks channels by verbosity.
    #
    #   0  ename     character TF-IDF, MEASURED 44% recovery at k=5, 2.6% precision
    #   1  nameonly  my dense version of the same idea, UNMEASURED - capped to its top 5 so
    #                it can add diversity without dominating
    #   2  e5_small tail  ranks 64-99, bounded at <=0.095 points by measurement
    PRIO = {"ename": 0, "nameonly": 1}
    parts = []
    for ch in RESCUE:
        d = load_channel(a.split, ch)
        if d is None:
            continue
        if ch != "ename" and a.rescue_rank_max:
            before = len(d)
            d = d[d["r"] < a.rescue_rank_max]
            print(f"    {ch}: kept {len(d):,} of {before:,} at rank < {a.rescue_rank_max}")
        d = d.copy()
        d["src"] = ch
        d["prio"] = PRIO.get(ch, 1)
        parts.append(d[["q", "c", "s", "r", "src", "prio"]])

    wide = load_channel(a.split, "e5_small")
    if wide is not None:
        w = wide[wide["r"] >= a.wide_rank_min].copy()
        print(f"  e5_small   kept {len(w):,} at rank >= {a.wide_rank_min}")
        w["src"] = "e5_small_tail"
        w["prio"] = 2
        parts.append(w[["q", "c", "s", "r", "src", "prio"]])

    if not parts:
        sys.exit("no rescue channels found - run 08_embed.py and 03_retrieve.py first")

    cand = pd.concat(parts, ignore_index=True)
    cand = cand.sort_values(["prio", "s"], ascending=[True, False]) \
               .drop_duplicates(["q", "c"], keep="first")
    print(f"\nunion of rescue channels: {len(cand):,} unique pairs")

    key = list(map(tuple, cand[["q", "c"]].to_numpy().tolist()))
    fresh = np.fromiter((k not in seen for k in key), bool, len(key))
    cand = cand[fresh]
    print(f"  not already scored by model-4: {len(cand):,} ({fresh.mean():.1%})")
    # These enter as a GUARANTEED UNION, never through rank fusion. They are address-less
    # records the other retrievers rank low; under RRF with one channel voting they land
    # below the K cut and the whole channel is thrown away. Priority 0 keeps them.

    cand = cand.sort_values(["q", "prio", "s"], ascending=[True, True, False])
    cand = cand.groupby("q", sort=False).head(a.per_s1)
    print(f"  after per-S1 cap of {a.per_s1}: {len(cand):,} over {cand.q.nunique():,} S1")
    print(f"  by source: {cand.src.value_counts().to_dict()}")

    if len(cand) > a.max_total:
        cand = cand.sort_values(["prio", "s"], ascending=[True, False]).head(a.max_total)
        print(f"  TRIMMED to --max-total {a.max_total:,}")

    d = os.path.dirname(P.work("prep", a.split, "_"))
    s1 = pd.read_parquet(os.path.join(d, "source1.parquet"))
    pool = pd.read_parquet(os.path.join(d, "pool.parquet"))
    qi, ci = cand["q"].to_numpy(), cand["c"].to_numpy()
    qn, qa, qc = (s1[k].to_numpy() for k in ("business_name", "business_address", "country"))
    pn, pa, pc = (pool[k].to_numpy() for k in ("business_name", "business_address", "country"))

    work = pd.DataFrame({
        "q": qi.astype(np.int32), "c": ci.astype(np.int32),
        "text_a": [CE.pair_text(qn[i], qa[i], qc[i]) for i in qi],
        "text_b": [CE.pair_text(pn[i], pa[i], pc[i]) for i in ci],
    })
    work["len_a"] = work["text_a"].str.len().astype(np.int32)
    work["len_b"] = work["text_b"].str.len().astype(np.int32)

    rate = 2400 * a.nshards
    print(f"\n{len(work):,} pairs to score. At ~2,400 pairs/s per lane x {a.nshards} lanes "
          f"= {len(work)/rate/3600:.1f} h wall clock.")

    shards = F.split_shards(work, a.nshards)
    man = []
    for i, sh in enumerate(shards):
        p = P.work("shards", f"{a.split}_shard{i}of{a.nshards}.parquet")
        sh.to_parquet(p, index=False, compression="zstd")
        mb = os.path.getsize(p) / 2 ** 20
        man.append({"shard": i, "n": len(sh), "mb": round(mb, 1),
                    "file": os.path.basename(p)})
        print(f"  shard {i}: {len(sh):>9,} pairs  {mb:6.1f} MB")
    import json
    with open(P.work("shards", f"{a.split}_manifest.json"), "w") as fh:
        json.dump({"split": a.split, "nshards": a.nshards, "total": len(work),
                   "shards": man}, fh, indent=1)
    print(f"\nwrote {P.work('shards', f'{a.split}_manifest.json')}")


if __name__ == "__main__":
    main()