如何获得集合的所有子集? (电源集) [英] How to get all subsets of a set? (powerset)

问题描述

提供一套

{0, 1, 2, 3}

如何生成子集:

[set(),
 {0},
 {1},
 {2},
 {3},
 {0, 1},
 {0, 2},
 {0, 3},
 {1, 2},
 {1, 3},
 {2, 3},
 {0, 1, 2},
 {0, 1, 3},
 {0, 2, 3},
 {1, 2, 3},
 {0, 1, 2, 3}]

推荐答案

Python

The Python itertools page has exactly a powerset recipe for this:

from itertools import chain, combinations

def powerset(iterable):
    "powerset([1,2,3]) --> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)"
    s = list(iterable)
    return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))

输出:

>>> list(powerset("abcd"))
[(), ('a',), ('b',), ('c',), ('d',), ('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd'), ('a', 'b', 'c'), ('a', 'b', 'd'), ('a', 'c', 'd'), ('b', 'c', 'd'), ('a', 'b', 'c', 'd')]

如果您一开始不喜欢该空元组，则可以将range语句更改为range(1, len(s)+1)以避免长度为0的组合.

If you don't like that empty tuple at the beginning, you can just change the range statement to range(1, len(s)+1) to avoid a 0-length combination.

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