在列表中進行列表的所有可能組合

Question

我已經找到了解決這個問題的幾種方法，但這不是我打算做的。

當我有清單時：

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

我想擁有所有可能的組合：

[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]
....

但是在python中有一個簡單的解決方案嗎？

謝謝，也可以創建1個列表而不是3個列表，例如：[7,8,9,4,5,6,1,2,3]

Answer 1

我認為itertools.permutations是您要尋找的。

>>> 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])]

並合並在一起：

>>> [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])

使用reduce組合列表：

>>> 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中，您無需指定排列的長度

>>> 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])]