通过python形成数组的排列组合

itertools
itertools 模块下提供了一些用于生成排列组合的工具函数 2。

用法
方法 含义
product(p, q, … [repeat=1]) 用序列 p、q、…序列中的元素进行排列（元素会重复）。就相当于使用嵌套循环组合。
permutations(p[, r]) 从序列 p 中取出 r 个元素的组成全排列，组合得到元组作为新迭代器的元素。
combinations(p, r) 从序列 p 中取出 r 个元素组成全组合，元素不允许重复，组合得到元组作为新迭代器的元素。
combinations_with_replacement(p, r) 从序列 p 中取出 r 个元素组成全组合，元素允许重复，组合得到元组作为新迭代器的元素。

示例
以下面的列表为例，

import itertools

L = [1, 2, 3, 4]
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2
3

不考虑顺序
comb1 = list(itertools.combinations([1,2,3,4],1))
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[(1,), (2,), (3,), (4,)]
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comb2 = list(itertools.combinations([1,2,3,4],2))
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[(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]
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comb3 = list(itertools.combinations([1,2,3,4],3))
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[(1, 2, 3), (1, 2, 4), (1, 3, 4), (2, 3, 4)]
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考虑顺序
comb1 = list(itertools.permutations([1,2,3,4],1))
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[(1,), (2,), (3,), (4,)]
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comb2 = list(itertools.permutations([1,2,3,4],2))
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[(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)]
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comb3 = list(itertools.permutations([1,2,3,4],3))
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[(1, 2, 3), (1, 2, 4), (1, 3, 2), (1, 3, 4), (1, 4, 2), (1, 4, 3), (2, 1, 3), (2, 1, 4), (2, 3, 1), (2, 3, 4), (2, 4, 1), (2, 4, 3), (3, 1, 2), (3, 1, 4), (3, 2, 1), (3, 2, 4), (3, 4, 1), (3, 4, 2), (4, 1, 2), (4, 1, 3), (4, 2, 1), (4, 2, 3), (4, 3, 1), (4, 3, 2)]
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numpy
示例
以下面的列表为例，

import numpy as np
import random

L = [1, 2, 3, 4]
1
2
3
4

不考虑顺序
elecnt = len(L)
cnt = 2
A = np.product(np.arange(1, elecnt+1))
B = np.product(np.arange(1, elecnt+1-cnt)) * np.product(np.arange(1, 1+cnt))
combcnt = int(A/B)

def combinations(combcnt, elecnt, cnt):
trials = combcnt
rslt = []
while trials > 0:
sample = sorted(random.sample(list(range(1, 1+elecnt)), cnt))
if sample in rslt:
continue
else:
if len(rslt) == combcnt:
break
else:
rslt.append(sample)
trials -= 1
return sorted(rslt)

comb = combinations(combcnt=combcnt, elecnt=elecnt, cnt=cnt)
print(f"self-defined: #{len(comb)}: {comb}")
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self-defined: #6: [[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]]
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import itertools
comb = list(itertools.combinations(L, 2))
print(f’itertools : #{len(comb)}: {comb}‘)
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itertools : #6: [(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]
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利用上文介绍到的 itertools.combinations 的方法，检验自定义函数 combinations 的结果的准确程度。

考虑顺序
参考这里 3

array_L = np.array(L)
cnt = 3

comb = np.unique(np.array(np.meshgrid(array_L, array_L, array_L)).T.reshape(-1, cnt), axis=0)
print(f"\ncomb:\n{comb}")
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[[1 1 1]
[1 2 1]
[1 3 1]
[1 4 1]
[2 1 1]
[2 2 1]
…
[4 1 4]
[4 2 4]
[4 3 4]
[4 4 4]]

shape, (64, 3)

‘’’

import itertools

comb = list(itertools.product(L, repeat=cnt))
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[[1 1 1]
[1 2 1]
[1 3 1]
[1 4 1]
[2 1 1]
[2 2 1]
…
[4 1 4]
[4 2 4]
[4 3 4]
[4 4 4]]

shape, (64, 3)

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