从 Python 中的列表中查找唯一的三胞胎

Question

我正在 python 中编写一些研究脚本，我有一个包含 9 个元素的列表：

a = [1,2,3,4,5,6,7,8,9]

我想找到包含 3 个元素（而不是 3 个元素集本身）的子集的所有组合，例如：

a_new = [ [ [1,2,3],[4,5,6],[7,8,9] ] ],
          [ [2,3,4],[5,6,7],[8,9,1] ],          
          [ [3,4,6],[1,2,8],[5,7,9] ], etc.... ]

元素在子集中不重复
集合 [3,2,1] 和 [1,2,3] 是同一集合并且不唯一

您是否遇到过这种情况，或者某些 Python 图书馆会处理它？

问候

Answer 1

你可以使用itertools内置模块

from itertools import combinations

a = [1,2,3,4,5,6,7,8,9]

combination = list(combinations(a, 3))

此输出为元组，您可以轻松转换为列表。

Answer 2

使用itertools.combinations如下：

from itertools import combinations

a = {1, 2, 3, 4, 5, 6, 7, 8, 9}

# pick 3
for c_out in combinations(a, r=3):
    s_out = set(c_out)
    # pick the next 3 from the those elements that are not already picked (i.e not in s_out)
    for c_in in combinations(a - s_out, r=3):
        # there only 3 elements left so only one way to pick
        rest = tuple(a - (s_out | set(c_in)))
        print(c_out, c_in, rest)

输出 （部分）

(1, 2, 3) (4, 5, 6) (8, 9, 7)
(1, 2, 3) (4, 5, 7) (8, 9, 6)
(1, 2, 3) (4, 5, 8) (9, 6, 7)
(1, 2, 3) (4, 5, 9) (8, 6, 7)
(1, 2, 3) (4, 6, 7) (8, 9, 5)
(1, 2, 3) (4, 6, 8) (9, 5, 7)
(1, 2, 3) (4, 6, 9) (8, 5, 7)
(1, 2, 3) (4, 7, 8) (9, 5, 6)
(1, 2, 3) (4, 7, 9) (8, 5, 6)
(1, 2, 3) (4, 8, 9) (5, 6, 7)
(1, 2, 3) (5, 6, 7) (8, 9, 4)
(1, 2, 3) (5, 6, 8) (9, 4, 7)
(1, 2, 3) (5, 6, 9) (8, 4, 7)
(1, 2, 3) (5, 7, 8) (9, 4, 6)

您可以在 documentation.

中阅读更多关于集合操作（并集、差集）的信息

从 Python 中的列表中查找唯一的三胞胎

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