如何查找列表列表中的哪些项目等于另一个列表 [英] How to find which items in list of lists is equal to another list

问题描述

我有一个看起来像这样的列表列表:

I have a list of lists that looks like this:

[[0],
[0, 1, 2],
[2],
[3],
[4],
[5],
[0, 1, 2, 3, 4, 5, 6, 7],
[7],
[8],
[9],
[8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18],
[11],
[11, 12, 13, 14, 15, 16, 17, 18],
[13],
[14],
[14, 15, 16, 17, 18],
[16, 17, 18],
[17],
[17, 18]]

我试图在连接时找到列表中最少的项目，这些项目等于列表的整个范围.在这种情况下，列表的整个范围是这样:

I am trying to find the least number of items in the list, when concatenated, that equal the full range of the list. In this case, the full range of the list is this:

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]

因此，在这种情况下，列表列表中的这两项将等于整个范围:

So in this case, these two items from the list of lists would equal the full range:

[0]
[0, 1, 2]
[2]
[3]
[4]
[5]
---> [0, 1, 2, 3, 4, 5, 6, 7]
[7]
[8]
[9]
---> [8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]
[11]
[11, 12, 13, 14, 15, 16, 17, 18]
[13]
[14]
[14, 15, 16, 17, 18]
[16, 17, 18]
[17]
[17, 18]

推荐答案

使用 itertools.permutations 和 chain 的一种方法:

from itertools import permutations, chain

starget = sorted(target)
for i in range(2, len(target)):
    for perm in permutations(l, i):
        if sorted(chain(*perm)) == starget:
            print(i, perm)
            break
    break

输出:

2 ([0, 1, 2, 3, 4, 5, 6, 7], [8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18])

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