Python中两个嵌套列表的交集 [英] Intersection of two nested lists in Python

问题描述

我对嵌套列表有疑问.我想用python语言计算两个嵌套列表的交集的长度.我的列表组成如下:

I've a problem with the nested lists. I want to compute the lenght of the intersection of two nested lists with the python language. My lists are composed as follows:

list1 = [[1,2], [2,3], [3,4]]
list2 = [[1,2], [6,7], [4,5]]
output_list = [[1,2]]

如何计算两个列表的交集?

How can i compute the intersection of the two lists?

推荐答案

我认为有两种合理的方法可以解决此问题.

I think there are two reasonable approaches to solving this issue.

如果顶层列表中没有太多项目，则可以简单地检查其中一个子列表中的每个子列表是否存在:

If you don't have very many items in your top level lists, you can simply check if each sub-list in one of them is present in the other:

intersection = [inner_list for inner in list1 if inner_list in list2]

in运算符将测试是否相等，因此可以按预期找到内容相同的不同列表对象.但是，这不是很有效，因为列表成员资格测试必须遍历所有子列表.换句话说，其性能为O(len(list1)*len(list2)).但是，如果您的列表很长，则可能要花费比您想要的更多的时间.

The in operator will test for equality, so different list objects with the same contents be found as expected. This is not very efficient however, since a list membership test has to iterate over all of the sublists. In other words, its performance is O(len(list1)*len(list2)). If your lists are long however, it may take more time than you want it to.

一种渐近有效的替代方法是将内部列表转换为tuple s并将顶级list s转换为set s.您实际上不需要为此编写任何循环，因为map和set类型的&运算符将为您处理所有这些事情:

A more asymptotically efficient alternative approach is to convert the inner lists to tuples and turn the top level lists into sets. You don't actually need to write any loops yourself for this, as map and the set type's & operator will take care of it all for you:

intersection_set = set(map(tuple, list1)) & set(map(tuple, list2))

如果需要将结果作为list的list，则可以将tuple s的set转换回list s的list:

If you need your result to be a list of lists, you can of course, convert the set of tuples back into a list of lists:

intersection_list = list(map(list, intersection_set))

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