迭代适合特定键的排列
想象一下列表
L = [(1,2,3),(4,5,6),(7,8),(9,10),(11,12,13)]
我想遍历这个列表的所有排列，这些排列适合任意长度的关键字，如(2,2,3,3,3) 。
所以在这种情况下，元素的长度适合该键的所有排列。
[(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)]
等等
现在，我只是遍历所有的排列组合，只是采用适合关键字的排列，但是这经历了很多浪费时间和排列。 尽管更深入地研究itertools，我宁愿直接生成我所需要的东西，但我不知道如何去做。
您可以精确地构建不同长度的元组的排列，并将它们组合在一起：
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键：
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]