如何根据另一个列表保留列表中的元素 [英] how to keep elements of a list based on another list

问题描述

我有两个看起来像这样的列表:

I have two lists looking like:

list1 = ['a','a','b','b','b','c','d','e','e','g','g']

list2 = ['a','c','z','y']

我想要做的是将list1的所有那些元素都保留在list2中. 结果应该是:

What I want to do is to keep all those elements of list1 that are also in list2. the outcome should be:

outcome= ['a','a','c']

推荐答案

使用in运算符，可以检查元素是否在序列中.

Using in operator, you can check whether an element is in a seqeunce.

>>> list2 = ['a','c','z','y']
>>> 'x' in list2
False
>>> 'y' in list2
True

使用列表理解:

>>> list1 = ['a','a','b','b','b','c','d','e','e','g','g']
>>> list2 = ['a','c','z','y']
>>> [x for x in list1 if x in list2]
['a', 'a', 'c']

但是x in list效率不高.最好将list2转换为 set 对象.

But x in list is not efficient. You'd better convert list2 to a set object.

>>> set2 = set(list2)
>>> [x for x in list1 if x in set2]
['a', 'a', 'c']

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