python re.sub() with pair or characters

I am trying to grab a string that is surrounded by "! and replace the first "!" with a "!_".

For example: str(!test!).strip() -> str(!_test!).strip()

Here is the code I have so far:

print re.sub(r'!.*?!','!_', 'str(!test!).strip()')

With this code I grab too much and the result is: str(!_).strip()

Any thoughts on how to zero in on the first "!". Or alternatively, is there a way to grab the string in the "!!" and then add "!_"+"!" arr开发者_运维技巧ound that string?

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Exclude ! from the characters between the !s: use [^!] instead of .

Then, capture the part of the RE you want to keep with (), and in the replacement string, use \1 to insert it again.

print re.sub(r'!([^!]*!)', r'!_\1', 'str(!test!).strip()')

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print re.sub(r'!(?=.*?!)', '!_', 'str(!test!).strip()')

Uses a positive lookahead.

print re.sub(r'!(.*?)!', r'!_\1!', 'str(!test!).strip()')

Uses a backreference.

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You need to group the second half of the "word" with parenthesis and use the group in your substitute string \g<1>.

re.sub(r'!(.*?!)', '!_\g<1>', 'str(!test!).strip()')