列表推导:如何为x的每次循环迭代创建一个新的列表，其中两个“列表"中的一项都被替换.比赛? [英] List Comprehensions: How do I crank out a new list for every loop iteration of x where an item in both "lists" match?

问题描述

allDasTickets = ["9255955", "9255958", "9255960", "9255977"]
[j for j in allDasTickets for x in allDasTickets if x != j]

因此，我得到了以下列表的列表:

So I get an list of the following lists:

['9255958', '9255960', '9255977']
['9255955', '9255958', '9255977']
['9255958', '9255960', '9255977']
['9255955', '9255958', '9255960']

即缺少匹配项的列表(我希望这是我上面提到的内容)

i.e. a list with the matching one missing (I hope that's what I've put above)

推荐答案

您忘记了内括号.之所以需要它们，是因为您试图生成一个列表列表，而不只是一个平面列表.

You forgot the inner brackets. You need them because you're trying to generate a list of lists, instead of just a flat list.

allDasTickets = ["9255955", "9255958", "9255960", "9255977"]
[[j for j in allDasTickets if x != j] for x in allDasTickets]

收益

[['9255958', '9255960', '9255977'],
 ['9255955', '9255960', '9255977'],
 ['9255955', '9255958', '9255977'],
 ['9255955', '9255958', '9255960']]

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