找到选择的最佳重新分区

Question

我有几个人在做x选择，在n可能性中按优先顺序排列。每个可能性只能分配一次

我想找到问题的所有解决方案，这样每个人都可以选择最低级别xmin

例如，对于x=3和说n=20，10个人在做选择：

    g1 = (3, 10, 11)  # g1 makes choices 3, 10 and 11 in order of preference
    g2 = (10, 9, 5)
    g3 = (10, 15, 3)
    g4 = (5, 9, 14)
    g5 = (10, 3, 7)
    ...
    g10 = (4, 19, 2)

在Python中，如何编写问题以找到解决方案，以便所有人都有一个至少分配了级别2（xmin=2）的选择？如果xmin=2没有解决方案，那么第3级（xmin=3）

我认为这与itertools有关，但我对这个问题不太清楚

编辑：再仔细考虑一下这个问题，我想到了这样一个问题：

import itertools

xmin = 2

groups = [g1, g2, g3, g4, g5]
sample = [g[:xmin] for g in groups]

[seq for seq in itertools.product(*sample) if len(seq) == len(set(seq))]

我没有醒，答案其实很简单

Answer 1

最后，itertools一如既往地简化了生活：）

使用此数据集：

gr1 = (3, 10, 11)
gr2 = (10, 9, 5)
gr3 = (10, 15, 3)
gr4 = (5, 9, 14)
gr5 = (10, 22, 7)
gr6 = (24, 9, 17)
gr7 = (10, 17, 3)
gr8 = (14, 21, 2)
gr9 = (14, 12, 21)
gr10 = (9, 1, 22)
gr11 = (19, 2, 8)
gr12 = (3, 8, 24)
gr13 = (9, 5, 13)
gr14 = (9, 1, 12)

以下是问题的解决方案：

import itertools
import numpy as np

xmin = 3
groups = [gr1, gr2, gr3, gr4, gr5, gr6, gr7, gr8, gr9, gr10, gr11, gr12, gr13, gr14]


sample = [g[:xmin] for g in groups]
assign = [seq for seq in itertools.product(*sample) if len(seq) == len(set(seq))]
choices = [[sample[i].index(assign[j][i]) for i in range(len(sample))] for j in range(len(assign))]

print(f'{len(choices)} possibilities found')

pertinence = [sum(choices[i]) for i in range(len(choices))]
best = np.array(assign)[np.array(pertinence) == min(pertinence)]

print(f'{len(best)} best choices:')
print(best)         

如果有人有更优雅的解决方案