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，其中列表中的列表数量不确定：

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]]
         ]

我现在无法弄清楚是如何生成所有可能组合的排列，同时保持第一级lists的顺序相同。我已经搞乱了嵌套的for循环和any()函数，但收效甚微。嵌套for循环不起作用，因为实际上len(lists)要大得多，并且需要len(lists)个for个循环。有没有人有任何想法？

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建议的那样，您可能正在寻找笛卡儿产品。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])]

如何在维护Python最高级别的顺序的同时生成列表列表的排列？

1 个答案: