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使用 pythons itertools,我想在一堆列表的所有排列的外积上创建一个迭代器。一个明确的例子:

import itertools
A = [1,2,3]
B = [4,5]
C = [6,7]

for x in itertools.product(itertools.permutations(A),itertools.permutations(B),itertools.permutations(C)):
    print x

虽然这可行,但我想将其推广到任意列表列表。我试过:

for x in itertools.product(map(itertools.permutations,[A,B,C])):
    print x

但它没有达到我的预期。预期的输出是:

((1, 2, 3), (4, 5), (6, 7))
((1, 2, 3), (4, 5), (7, 6))
((1, 2, 3), (5, 4), (6, 7))
((1, 2, 3), (5, 4), (7, 6))
((1, 3, 2), (4, 5), (6, 7))
((1, 3, 2), (4, 5), (7, 6))
((1, 3, 2), (5, 4), (6, 7))
((1, 3, 2), (5, 4), (7, 6))
((2, 1, 3), (4, 5), (6, 7))
((2, 1, 3), (4, 5), (7, 6))
((2, 1, 3), (5, 4), (6, 7))
((2, 1, 3), (5, 4), (7, 6))
((2, 3, 1), (4, 5), (6, 7))
((2, 3, 1), (4, 5), (7, 6))
((2, 3, 1), (5, 4), (6, 7))
((2, 3, 1), (5, 4), (7, 6))
((3, 1, 2), (4, 5), (6, 7))
((3, 1, 2), (4, 5), (7, 6))
((3, 1, 2), (5, 4), (6, 7))
((3, 1, 2), (5, 4), (7, 6))
((3, 2, 1), (4, 5), (6, 7))
((3, 2, 1), (4, 5), (7, 6))
((3, 2, 1), (5, 4), (6, 7))
((3, 2, 1), (5, 4), (7, 6))
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1 回答 1

12

您错过了*将列表解压缩为 3 个参数的方法

itertools.product(*map(itertools.permutations,[A,B,C]))
于 2012-08-22T00:17:11.563 回答