Question

想象一下列表

L = [(1,2,3),(4,5,6),(7,8),(9,10),(11,12,13)]

我想遍历此列表中适合任意长度键的所有排列，例如(2,2,3,3,3)。

所以在这种情况下，所有元素的长度都适合该键的排列。

[(7,8),(9,10),(1,2,3),(4,5,6),(11,12,13)]

[(9,10),(7,8),(1,2,3),(4,5,6),(11,12,13)]

[(7,8),(9,10),(4,5,6),(1,2,3),(11,12,13)]

等

现在我只是迭代所有排列，只采用适合键的那些排列，但这会耗费大量浪费的时间和排列。我宁愿直接生成我需要的东西，但我根本不知道该怎么做，尽管深入研究了itertools。

Answer 1

你可以组织不同长度的元组的排列并将它们组合起来：

from itertools import chain, permutations, product
tuples_by_length = {}
for t in L:
    tuples_by_length.setdefault(len(t), []).append(t)
for x, y in product(permutations(tuples_by_length[2]),
                    permutations(tuples_by_length[3])):
    print list(chain(x, y))

输出：

[(7, 8), (9, 10), (1, 2, 3), (4, 5, 6), (11, 12, 13)]
[(7, 8), (9, 10), (1, 2, 3), (11, 12, 13), (4, 5, 6)]
[(7, 8), (9, 10), (4, 5, 6), (1, 2, 3), (11, 12, 13)]
[(7, 8), (9, 10), (4, 5, 6), (11, 12, 13), (1, 2, 3)]
[(7, 8), (9, 10), (11, 12, 13), (1, 2, 3), (4, 5, 6)]
[(7, 8), (9, 10), (11, 12, 13), (4, 5, 6), (1, 2, 3)]
[(9, 10), (7, 8), (1, 2, 3), (4, 5, 6), (11, 12, 13)]
[(9, 10), (7, 8), (1, 2, 3), (11, 12, 13), (4, 5, 6)]
[(9, 10), (7, 8), (4, 5, 6), (1, 2, 3), (11, 12, 13)]
[(9, 10), (7, 8), (4, 5, 6), (11, 12, 13), (1, 2, 3)]
[(9, 10), (7, 8), (11, 12, 13), (1, 2, 3), (4, 5, 6)]
[(9, 10), (7, 8), (11, 12, 13), (4, 5, 6), (1, 2, 3)]

这种方法可以推广到任意长度的密钥：

def permutations_with_length_key(lst, length_key):
    tuples_by_length = {}
    for t in L:
        tuples_by_length.setdefault(len(t), []).append(t)
    positions = {k: i for i, k in enumerate(tuples_by_length.iterkeys())}
    for x in product(*[permutations(v) for v in tuples_by_length.itervalues()]):
        x = map(iter, x)
        yield [next(x[positions[y]]) for y in length_key]

迭代适合特定键的排列

1 个答案: