如何用python re.sub仅替换部分匹配 [英] How to replace only part of the match with python re.sub

问题描述

我需要用一个 reg 表达式匹配两个 case 并进行替换

I need to match two cases by one reg expression and do replacement

'long.file.name.jpg' -> 'long.file.name_suff.jpg'

'long.file.name.jpg' -> 'long.file.name_suff.jpg'

'long.file.name_a.jpg' -> 'long.file.name_suff.jpg'

'long.file.name_a.jpg' -> 'long.file.name_suff.jpg'

我正在尝试执行以下操作

I'm trying to do the following

re.sub('(\_a)?.[^.]*$' , '_suff.',"long.file.name.jpg")

但这是删除扩展名.jpg"，我得到了

But this is cut the extension '.jpg' and I'm getting

long.file.name_suff.而不是 long.file.name_suff.jpg我知道这是因为 [^.]*$ 部分，但我不能排除它，因为我必须找到最后一次出现的 '_a' 来替换或最后一次 '.'

long.file.name_suff. instead of long.file.name_suff.jpg I understand that this is because of [^.]*$ part, but I can't exclude it, because I have to find last occurance of '_a' to replace or last '.'

有没有办法只替换匹配的一部分?

Is there a way to replace only part of the match?

推荐答案

 re.sub(r'(?:_a)?.([^.]*)$', r'_suff.1', "long.file.name.jpg")

?: 开始一个不匹配的组(SO 答案)，所以 (?:_a) 匹配 _a 但没有枚举它，下面的问号使它成为可选的.

?: starts a non matching group (SO answer), so (?:_a) is matching the _a but not enumerating it, the following question mark makes it optional.

所以在英语中，这表示，匹配结尾 .<anything> 跟随(或不跟随)模式 _a

So in English, this says, match the ending .<anything> that follows (or doesn't) the pattern _a

另一种方法是使用lookbehind(见这里).提到这个是因为它们非常有用，但我做了 15 年的 RE 才知道它们

Another way to do this would be to use a lookbehind (see here). Mentioning this because they're super useful, but I didn't know of them for 15 years of doing REs

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