最快的方法是在矩阵迭代和排除的矩阵中找到最高价值的索引

Question

矩阵

sim = np.array([[1,5,6,2],[7,10,3,4],[1,5,3,7]])
pairwise_best_hit(sim)

returns[[1, 2, 0], [1, 3, 2]]

这里是仅使用numpy的候选人：

import numpy as np
import pandas as pd
import timeit

sim = np.array([[1,5,6,2],[7,10,3,4],[1,5,3,7]])

def mine(sim):
    out = []
    copy = sim.copy()
    MIN = np.iinfo(copy.dtype).min
    for _ in range(min(copy.shape)):
        ij = np.unravel_index(copy.argmax(), copy.shape)
        out.append(ij)
        copy[ij[0]] = MIN
        copy[:,ij[1]] = MIN
    return np.transpose(out)

def yours(sim):
    xdim,ydim = np.meshgrid(np.arange(sim.shape[1]),np.arange(sim.shape[0]))
    table = np.vstack((sim.ravel(),xdim.ravel(),ydim.ravel())).T
    df = pd.DataFrame(table).rename(columns={0:'sim',1:'index2',2:'index1'}).sort_values('sim',ascending=False)
    seq1_hits = []
    seq2_hits = []
    while len(df):
        index1 = df.iloc[0]['index1']
        index2 = df.iloc[0]['index2']
        seq1_hits.append(index1)
        seq2_hits.append(index2)
        df = df[(df['index1']!=index1)&(df['index2']!=index2)]
    return [seq1_hits,seq2_hits]

assert np.all(mine(sim) == yours(sim))

%timeit yours(sim)
# 1.05 ms ± 6.78 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)

%timeit mine(sim)
# 8.18 µs ± 19.4 ns per loop (mean ± std. dev. of 7 runs, 100,000 loops each)

Answer 1