以子列表的形式获取列表的所有可能组合 [英] getting all possible combinations of a list in a form of sublists

问题描述

我想知道是否有人可以帮助完成以下任务：
当顺序无关紧要时，可以将列表的所有组合拆分为子列表的方法是什么？

I wonder if someone can help with the following task: What is the way to get all combinations a list can be split into sublists, when order does not matter?

假设我有4个项目的列表：

Let's say I have a list of 4 items:

import itertools as it

a = [1, 2, 3, 4]
print(list(it.combinations(a, 2)))

这将给我列出6对可能的货币对：

That will give me a list of 6 possible pairs:

[(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]

如何（以其他方式或其他方式）制作一组包含原始 [1、2、3、4] 以任何顺序排列？因此，对于此示例，它将包含三个子列表：

How to make (out of it? or any other way) a set of lists that contain original [1, 2, 3, 4] sequence in any order? So for this example it will contain three sublists:

 [(1, 2), (3, 4)]
 [(1, 3), (2, 4)]
 [(1, 4), (2, 3)]

更新：
一个小澄清：
换句话说，我需要获取所有n元组，以便其成员包含原始列表的所有填充，当一个n元组中的顺序无关紧要时。因此 [（1，2），（3，4）] 可以，但是 [（2，1），（3，4）] 不需要，因为它与第一个集合相同，如果我们忽略该顺序。

UPDATE: A small clarification: In other words, I need to get all sets of n-tuples such that their members contain all the population of original list, when the order within an n-tuple does not matter. Thus [(1, 2), (3, 4)] is ok, but [(2, 1), (3, 4)] is not needed because it is the same as the first set, if we ignore the order.

UPDATE2：
长度为6的列表，对于大小为2的块，此 fun 函数应如下所示：

UPDATE2: So for the list of length 6, and for chunks of size 2 this fun function should work as follows:

import itertools as it
a = [1, 2, 3, 4, 5, 6,]
r = 2

# fun(a,r):
# OUT:
# [
#    (1, 2), (3, 4), (5, 6)
#    (1, 3), (2, 4), (5, 6),
#    (1, 4), (2, 3), (5, 6),
#    (1, 5), (2, 3), (4, 6),
#    (1, 6), (2, 3), (4, 5),
#  ]

推荐答案

只需 zip 组合，并带有相反的组合，并且仅取前一半

Just zip the combinations, with its reverse and take only the first half of the resulting list

>>> import itertools as it
>>> lst = [1, 2, 3, 4]
>>> r = len(lst)//2
>>> combs = list(it.combinations(lst, r))
>>> list(it.islice(zip(combs, reversed(combs)), len(combs)//2))
[((1, 2), (3, 4)), ((1, 3), (2, 4)), ((1, 4), (2, 3))]

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