如何获得集合的所有子集? (电源集)

本文介绍了如何获得集合的所有子集? (电源集)的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 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 itertools页面具有与此完全相同的powerset配方:

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.