如何生成列表列表的排列，同时在 Python 中保持最高级别的顺序？

Question

我正在寻找一种简单的方法来解决这个问题。假设我有一个列表列表，其中有不确定数量的列表：

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

我现在想不通的是生成所有可能组合的排列，同时保持 lists 的第一级顺序相同。我搞砸了嵌套的 for 循环和 any() 函数，但收效甚微。嵌套的 for 循环不起作用，因为实际上 len(lists) 大得多，并且需要 len(lists) 个 for 循环。有人有什么想法吗？

在上面的例子中，一些可能的排列是：

[[1,2,3,4],
 [1,2,3,4],
 [1,2,3,4],
 [1,2,3,4]]

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

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

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

Answer 1

正如@DSM 所暗示的，您可能正在寻找笛卡尔积。排列意味着不同的东西。

>>> import pprint, itertools as it
>>> lists = [
...          [[1,2,3,4],[2,3,4,5]],
...          [[1,2,3,4],[2,3,4,5],[3,4,5,6]],
...          [[1,2,3,4]],
...          [[1,2,3,4],[2,3,4,5]]
...          ]
>>> pprint.pprint(list(it.product(*lists)))
[([1, 2, 3, 4], [1, 2, 3, 4], [1, 2, 3, 4], [1, 2, 3, 4]),
 ([1, 2, 3, 4], [1, 2, 3, 4], [1, 2, 3, 4], [2, 3, 4, 5]),
 ([1, 2, 3, 4], [2, 3, 4, 5], [1, 2, 3, 4], [1, 2, 3, 4]),
 ([1, 2, 3, 4], [2, 3, 4, 5], [1, 2, 3, 4], [2, 3, 4, 5]),
 ([1, 2, 3, 4], [3, 4, 5, 6], [1, 2, 3, 4], [1, 2, 3, 4]),
 ([1, 2, 3, 4], [3, 4, 5, 6], [1, 2, 3, 4], [2, 3, 4, 5]),
 ([2, 3, 4, 5], [1, 2, 3, 4], [1, 2, 3, 4], [1, 2, 3, 4]),
 ([2, 3, 4, 5], [1, 2, 3, 4], [1, 2, 3, 4], [2, 3, 4, 5]),
 ([2, 3, 4, 5], [2, 3, 4, 5], [1, 2, 3, 4], [1, 2, 3, 4]),
 ([2, 3, 4, 5], [2, 3, 4, 5], [1, 2, 3, 4], [2, 3, 4, 5]),
 ([2, 3, 4, 5], [3, 4, 5, 6], [1, 2, 3, 4], [1, 2, 3, 4]),
 ([2, 3, 4, 5], [3, 4, 5, 6], [1, 2, 3, 4], [2, 3, 4, 5])]

如何生成列表列表的排列，同时在 Python 中保持最高级别的顺序？

How can I generate a permutation of a list of lists while maintaining order on the highest level in Python?

python

list

permutation