How to replace only part of the match with python re.sub
PythonRegexPython Problem Overview
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_a.jpg' -> 'long.file.name_suff.jpg'
I'm trying to do the following
re.sub('(\_a)?\.[^\.]*$' , '_suff.',"long.file.name.jpg")
But this is cut the extension '.jpg' and I'm getting
long.file.name_suff. instead of long.file.name_suff.jpg I understand that this is because of [^.]*$ part, but I can't exclude it, because I have to find last occurance of '_a' to replace or last '.'
Is there a way to replace only part of the match?
Python Solutions
Solution 1 - Python
Put a capture group around the part that you want to preserve, and then include a reference to that capture group within your replacement text.
re.sub(r'(\_a)?\.([^\.]*)$' , r'_suff.\2',"long.file.name.jpg")
Solution 2 - Python
re.sub(r'(?:_a)?\.([^.]*)$', r'_suff.\1', "long.file.name.jpg")
?: starts a non matching group (SO answer), so (?:_a) is matching the _a but not enumerating it, the following question mark makes it optional.
So in English, this says, match the ending .<anything> that follows (or doesn't) the pattern _a
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
Solution 3 - Python
Just put the expression for the extension into a group, capture it and reference the match in the replacement:
re.sub(r'(?:_a)?(\.[^\.]*)$' , r'_suff\1',"long.file.name.jpg")
Additionally, using the non-capturing group (?:…) will prevent re to store to much unneeded information.
Solution 4 - Python
You can do it by excluding the parts from replacing. I mean, you can say to the regex module; "match with this pattern, but replace a piece of it".
re.sub(r'(?<=long.file.name)(\_a)?(?=\.([^\.]*)$)' , r'_suff',"long.file.name.jpg")
>>> 'long.file.name_suff.jpg'
long.file.name and .jpg parts are being used on matching, but they are excluding from replacing.
Solution 5 - Python
I wanted to use capture groups to replace a specific part of a string to help me parse it later. Consider the example below:
s= '<td> <address> 110 SOLANA ROAD, SUITE 102<br>PONTE VEDRA BEACH, FL32082 </address> </td>'
re.sub(r'(<address>\s.*?)(<br>)(.*?\<\/address>)', r'\1 -- \3', s)
##'<td> <address> 110 SOLANA ROAD, SUITE 102 -- PONTE VEDRA BEACH, FL32082 </address> </td>'
Solution 6 - Python
print(re.sub('name(_a)?','name_suff','long.file.name_a.jpg'))
# long.file.name_suff.jpg
print(re.sub('name(_a)?','name_suff','long.file.name.jpg'))
# long.file.name_suff.jpg