生成列表字典的所有排列的组合(按特定顺序)

Question

我有一份清单

d = {'A': [1,2,3], 'B': [4,5], 'C': [6]}

我需要生成每个列表(A,B和C)的所有排列.还行吧.

p = {}
for k in d.keys():
    p[k] = [i for i in itertools.permutations(d[k])]

这导致了 p

{'A': [(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)],
 'B': [(4, 5), (5, 4)],
 'C': [(6,)]}

然后我需要合并来自A,B和C列表的元组,但是按特定的顺序(例如按照sorted(p.keys())实际给出的顺序['A', 'B', 'C']).所以我应该获得整数元组的列表:

[(1,2,3,4,5,6),
 (1,2,3,5,4,6),
 (1,3,2,4,5,6),
 (1,3,2,5,4,6),
 ...
 (3,2,1,5,4,6)
]

我知道itertools.product可以在这种情况下使用,但是初始字典d可以包含具有不同键的任意数量的值,并且在这种情况下我不知道如何使用它.或许你可以建议完全不同的解决方案.最终解决方案运行得越快越好.

Answer 1

像这样的东西:

from itertools import permutations, product, chain

d = {'A': [1,2,3], 'B': [4,5], 'C': [6]}
# You don't need to materialise permutations here, but this matches your existing dict
p = {k:list(permutations(v)) for k, v in d.iteritems()}    

for blah in product(*map(p.get, sorted(p))):
    print list(chain.from_iterable(blah)) # or use tuple instead of list

[1, 2, 3, 4, 5, 6]
[1, 2, 3, 5, 4, 6]
[1, 3, 2, 4, 5, 6]
[1, 3, 2, 5, 4, 6]
[2, 1, 3, 4, 5, 6]
[2, 1, 3, 5, 4, 6]
[2, 3, 1, 4, 5, 6]
[2, 3, 1, 5, 4, 6]
[3, 1, 2, 4, 5, 6]
[3, 1, 2, 5, 4, 6]
[3, 2, 1, 4, 5, 6]
[3, 2, 1, 5, 4, 6]