给定一个集合
{0, 1, 2, 3}
我怎样才能产生子集:
[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}]
Pythonitertools页面正好有一个powerset秘诀:
itertools
powerset
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')]
如果您不喜欢开头的那个空元组,您可以更改range语句range(1, len(s)+1)以避免 0 长度组合。
range
range(1, len(s)+1)