How to replace into a regex group in Python

Question

>>> s = 'foo: "apples", bar: "oranges"' >>> pattern = 'foo: "(.*)"' 

I want to be able to substitute in the group:

 >>> re.sub(pattern, 'pears', s, group=1) 'foo: "pears", bar: "oranges"' 

Is there a good way to do this?

Answer 1

For me, something like works:

 rx = re.compile(r'(foo: ")(.*?)(".*)') s_new = rx.sub(r'\g<1>pears\g<3>', s) print(s_new) 

Pay attention to ? in re, so it ends first, " also pay attention to " in groups 1 and 3, because they must be output.

Instead of \g<1> (or \g<number> ) you can use only \1 , but do not forget to use raw strings and that the form g<1> preferable because \1 can be ambiguous (look for examples in Python doc ).

Answer 2

 re.sub(r'(?<=foo: ")[^"]+(?=")', 'pears', s) 

A regular expression matches a sequence of characters that

• Fulfills the line foo: " ,
• does not contain double quotes and Per
• follows "

(?<=) and (?=) are lookbehind and lookahead

This regular expression will not be executed if the foo value contains escaped quotas. Use the following command to catch them:

 re.sub(r'(?<=foo: ")(\\"|[^"])+(?=")', 'pears', s) 

Code example

 >>> s = 'foo: "apples \\\"and\\\" more apples", bar: "oranges"' >>> print s foo: "apples \"and\" more apples", bar: "oranges" >>> print re.sub(r'(?<=foo: ")(\\"|[^"])+(?=")', 'pears', s) foo: "pears", bar: "oranges"