如何使用仅包含项目的另一个列表来对item:value列表进行子集化? [英] How to subset an item:value list using another list with just items?

问题描述

我有2个列表.一个是单词及其频率的列表，另一个是单词的列表.

I have 2 lists. One is a list of words and their frequencies and the other is a list of words.

a = [('country',3478), ('island',2900),('river',5)] 
b = ['river','mountain','bank'] 

a中有成千上万的条目，而b中仅有数百.

There are thousands of entries in a but only hundreds in b.

如何子集列出a以便返回:

How can I subset list a so that i return:

c=[('river',5)]

给定条目数量，for循环将花费很长时间，我想列表理解是解决方案，但无法正确解决.

For loops would take too long given the number of entries and i imagine list comprehension is the solution but cannot get it right.

我的主要目标是使用最终列表创建wordcloud.任何帮助将不胜感激

My main goal is to then create a wordcloud with my final list. Any help would be appreciated

**编辑是因为某些评论者指出我犯了一个错误.我想返回

**Edited because I made a mistake as pointed out by some commenters. I want to return

c=[('river',5)]

代替

c=['river',5]

就像我最初写的那样.致歉，并感谢您指出

as i originally wrote. Apologies and thanks for pointing it out

推荐答案

我认为您实际上是想要的:

I assume you actually want:

c = [('river',5)]  # a list with one tuple

最好先在b中构造一组值:

You better first construct a set of values in b:

bd = set(b)

然后您可以使用列表理解:

then you can use list comprehension:

c = [(x,y) for x,y in a if x in bd]

话虽如此，如果您想查找单词的出现频率，我建议您不要构造一个元组列表，而应该构造一个字典.您可以通过字典理解

That being said, if you want to lookup the frequency of a word, I advice you not to construct a list of tuples, but a dictionary. You can do this with dictionary comprehension:

c = {x: y for x,y in a if x in bd}  # dictionary variant

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