在列表中列出所有可能的列表组合

Question

我发现了这个问题的几个答案，但它不是我想要做的。

当我有一个列表：

[1,2,3],[4,5,6],[7,8,9] 

我想有所有可能的组合：

[1,2,3],[7,8,9],[4,5,6] 
[1,2,3],[4,5,6],[7,8,9] 
[7,8,9],[4,5,6],[1,2,3] 
.... 

但有一个简单的解决方案，这在Python？

感谢了，是不是也可以创建1名列表，而不是像3： [7,8,9,4,5,6,1,2,3]

Answer 1

itertools.permutations是你在做什么寻找我的想法。

>>> import itertools 
>>> l = [1,2,3],[4,5,6],[7,8,9] 
>>> list(itertools.permutations(l, len(l))) 
[([1, 2, 3], [4, 5, 6], [7, 8, 9]), 
    ([1, 2, 3], [7, 8, 9], [4, 5, 6]), 
    ([4, 5, 6], [1, 2, 3], [7, 8, 9]), 
    ([4, 5, 6], [7, 8, 9], [1, 2, 3]), 
    ([7, 8, 9], [1, 2, 3], [4, 5, 6]), 
    ([7, 8, 9], [4, 5, 6], [1, 2, 3])] 

并合并在一起：

>>> [list(itertools.chain(*x)) for x in itertools.permutations(l, len(l))] 

[[1, 2, 3, 4, 5, 6, 7, 8, 9], 
[1, 2, 3, 7, 8, 9, 4, 5, 6], 
[4, 5, 6, 1, 2, 3, 7, 8, 9], 
[4, 5, 6, 7, 8, 9, 1, 2, 3], 
[7, 8, 9, 1, 2, 3, 4, 5, 6], 
[7, 8, 9, 4, 5, 6, 1, 2, 3]] 

Answer 2

>>> from itertools import permutations 
>>> a = [[1,2,3],[4,5,6],[7,8,9]] 
>>> for permu in permutations(a,3): 
... print permu 
... 
([1, 2, 3], [4, 5, 6], [7, 8, 9]) 
([1, 2, 3], [7, 8, 9], [4, 5, 6]) 
([4, 5, 6], [1, 2, 3], [7, 8, 9]) 
([4, 5, 6], [7, 8, 9], [1, 2, 3]) 
([7, 8, 9], [1, 2, 3], [4, 5, 6]) 
([7, 8, 9], [4, 5, 6], [1, 2, 3]) 

组合列表使用reduce：

>>> a = [[1,2,3],[4,5,6],[7,8,9]] 
>>> for permu in permutations(a,3): 
... print reduce(lambda x,y: x+y,permu,[]) 
... 
[1, 2, 3, 4, 5, 6, 7, 8, 9] 
[1, 2, 3, 7, 8, 9, 4, 5, 6] 
[4, 5, 6, 1, 2, 3, 7, 8, 9] 
[4, 5, 6, 7, 8, 9, 1, 2, 3] 
[7, 8, 9, 1, 2, 3, 4, 5, 6] 
[7, 8, 9, 4, 5, 6, 1, 2, 3] 

Answer 3

在Python2.7你不需要指定排列的长度

>>> T=[1,2,3],[4,5,6],[7,8,9] 
>>> from itertools import permutations 
>>> list(permutations(T)) 
[([1, 2, 3], [4, 5, 6], [7, 8, 9]), ([1, 2, 3], [7, 8, 9], [4, 5, 6]), ([4, 5, 6], [1, 2, 3], [7, 8, 9]), ([4, 5, 6], [7, 8, 9], [1, 2, 3]), ([7, 8, 9], [1, 2, 3], [4, 5, 6]), ([7, 8, 9], [4, 5, 6], [1, 2, 3])]