清单理解和交集问题 [英] List comprehension and intersection problem

问题描述

list(set(a[0]) & set(a[1]) & set(a[2]) & set(a[3]) & set(a[4]))

有人知道如何编写这种方式，这样我们就不必先验地知道我们将得到多少个列表? (即5个未硬编码)?

Does anyone know how to write this in a way such that we dont need to know apriori how many lists we will be given? (ie 5 not hard coded in)?

每个a都是大小不同的列表.

Each a is a list of varying size.

推荐答案

只要有至少一套，就可以这样做:

As long as you have at least one set, you can do this:

list(set(a[0]).intersection(*a[1:]))

如果可能没有集合，则必须自己决定无集合的交集"在您的应用程序中实际上意味着什么.如果要空集:

If there might be no sets, you have to decide for yourself what the "intersection of no sets" actually should mean in your application. If you want the empty set:

list(set(*a[:1]).intersection(*a[1:]))

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