如何在 Python 中的列表中为多个分组生成排列

Question

我知道我们可以使用 itertools.permutations 来排列列表中的不同项目，但是，如果我有一个列表，其中很少有项目需要处于固定位置，很少有项目需要与更多项目交换，并且需要有几个项目再换2个？

例如：

test = [1, 6, 2, 12, 5, 13, 11, 14, 15]

如何使用 Python itertools.permutation 或其他方法生成具有以下约束的所有可能组合？

更新：

1 and 5 have fixed positions
In position 2, I could have either 6 or 11
In position 3, I could have either 2 or 12
In position 4, I could have 2 or 12
In position 6, I could have either 13, 14, 15 and so on

所以，我的清单是这样的：

[1, (6, 11), (2, 12), (2,12), 5, (13, 14, 15), (6, 11), (13, 14, 15), (13, 14, 15)]

我已经将数字包含在组中，这表示同一组中的数字可以相互交换。

谢谢。

Answer 1

你可以这样做：

from itertools import permutations, product, chain

test = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]
groups = [[1], [2, 3, 4], [5], [6, 7], [8, 9], [10], [11, 12, 13], [14], [15, 16]]

result = [list(chain.from_iterable(permutation)) for permutation in product(*map(permutations, groups))]

for e in result[:20]:
    print(e)

输出

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 12, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 12, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 11, 13, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 11, 13, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 11, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 11, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 11, 12, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 11, 12, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 12, 11, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 12, 11, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 11, 12, 13, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 11, 12, 13, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 11, 13, 12, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 11, 13, 12, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 12, 11, 13, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 12, 11, 13, 14, 16, 15]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 12, 13, 11, 14, 15, 16]
[1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 12, 13, 11, 14, 16, 15]

更新

鉴于新的约束，您可以执行以下操作：

from functools import partial
from itertools import combinations, permutations, product, chain

choose_one = partial(lambda r, iterable: combinations(iterable, r), 1)
groups = [[[1]], combinations([6, 11], 1), permutations([2, 12]), [[5]], combinations([13, 14, 15], 1)]

for e in product(*groups, repeat=1):
    print(list(chain.from_iterable(e)))

输出

[1, 6, 2, 12, 5, 13]
[1, 6, 2, 12, 5, 14]
[1, 6, 2, 12, 5, 15]
[1, 6, 12, 2, 5, 13]
[1, 6, 12, 2, 5, 14]
[1, 6, 12, 2, 5, 15]
[1, 11, 2, 12, 5, 13]
[1, 11, 2, 12, 5, 14]
[1, 11, 2, 12, 5, 15]
[1, 11, 12, 2, 5, 13]
[1, 11, 12, 2, 5, 14]
[1, 11, 12, 2, 5, 15]