Python正则表达式为字符串中的所有单词添加一个字符，除了和

Question

I want to be able to generate 'foos, bars and bees' from 'foo, bar and bee' using re.sub.

I can't even get just adding 's' to all words to work. I'll work on excluding 'and' once I get that part. I have tried subbing \\b with "s" but that matches beginnings and endings of words. If I use '\\w*\\b' then the whole word is replaced. I am trying to figure this out using the Python docs, and (?P) or (?<=...) lookbehind assertions seem like they might be what I'm after, but I am having trouble getting those to cooperate, and the examples are limited.

Answer 1

This works, based on the replacement accepting a callable:

re.sub('(\w+)', lambda m: m.group(1) + 's' if m.group(1) != 'and' else 'and', 'foo, bar and bee')

It was inspired by an old bug report (second to last entry).

EDIT: A shorter and probably more readable solution:

re.sub('(and)|(\w+)', lambda m: m.group(1) or m.group(2) + 's', 'foo, bar and bee')

It also has the benefit of making it easier to add other words to the exception list, as isedev suggested in a comment.

Answer 2

Without considering words to exclude, the following will add an 's' to the end of all words in the string:

re.sub('([a-zA-Z]+)','\\1s','foo, bar and bee')
-> 'foos, bars ands bees'

To pluralise words in a more generic and less error prone way, you might want to take a look at the inflect package (for English at least).

Answer 3

The below code would add s to all the words except the word and ,

>>> import re
>>> s = "foo, bar and bee "
>>> m = re.sub(r'(?!and)(\b\w+\b)', r'\1s', s)
>>> m
'foos, bars and bees '

Negative lookahead asserts that it would match one or more word characters but not a \\band\\b . \\b here, means word boundary which matches between a word character and a non-word character.