按字符串中的单词数对字符串列表进行排序

本文介绍了按字符串中的单词数对字符串列表进行排序的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有一个这样的字符串列表:

I have a list of strings as such:

mylist = ["superduperlongstring", "a short string", "the middle"]

我想以最大单词数的字符串为首的方式进行排序，即

I want to sort this in such a way that the string with the largest number of words is first, ie,

mylist = ["a short string", "the middle", "superduperlongstring"]

这有点棘手，因为如果我按长度排序

Its a bit tricky, since if I sort in place by length

mylist.sort(key = len)

我回到了起点.

有没有人遇到过一个优雅的解决方案?谢谢.

Has anyone come across a graceful solution to this? Thanks.

推荐答案

假定单词由空格分隔，请调用 str.split 返回字符串包含的单词的列表:

Assuming that words are separated by whitespace, calling str.split with no arguments returns a list of the words that a string contains:

>>> "superduperlongstring".split() ['superduperlongstring'] >>> "a short string".split() ['a', 'short', 'string'] >>> "the middle".split() ['the', 'middle'] >>>

因此，您可以根据以下列表的长度对mylist进行排序，以获得所需的输出:

Therefore, you can get the output you want by sorting mylist based off of the length of these lists:

>>> mylist = ["superduperlongstring", "a short string", "the middle"] >>> mylist.sort(key=lambda x: len(x.split()), reverse=True) >>> mylist ['a short string', 'the middle', 'superduperlongstring'] >>>

您还需要将list.sort的reverse参数设置为True，如上所示.

You will also need to set the reverse parameter of list.sort to True, as shown above.