CODEWITHSUNDEEP
pythonregexstring
Shakesbeery
Shakesbeery

Reputation: 249

How to reference a regex match as a variable in python

So here's my problem:

I need to separate these punctuation items '], [, ?, !, (, ), ", ;, {, }' from whatever character they touch with a space. For example, 

"Did he eat it (the bug)?" becomes: " Did he eat it ( the bug ) ? "

I can do something like:

re.search(r'[]?!()";{}', mytext)

But when the search finds a match, how do I reference the item that was matched so I can replace it with itself and a space? In pseudo-code:

replace(matched_punc, matched_punc + " ")

Or the space could come before if it's word-final, but I can hash that out later. Mostly I just need to figure out how to replace something with itself and a space.

Many thanks.

Upvotes: 0

Views: 190

Answers (3)

p.s.w.g
p.s.w.g

Reputation: 149020

What about using re.sub:

re.sub(r'([][?!()";{}])', r' \1 ', mytext)

Or, if you need to ensure that you don't get multiple spaces in a together, something like should work:

re.sub(r'(?<=\S)(?=[][?!()";{}])|(?<=[][?!()";{}])(?=\S)', ' ', mytext)

Note: Thanks to perreal for making this click for me.

Upvotes: 3

perreal
perreal

Reputation: 97948

An alternative us to use lookaround expressions to do insertion instead of substitution:

print re.sub(r'(?<=[][?!()"])(?=[^ ])', ' ', 
        re.sub(r'(?<=[^ ])(?=[\[\]?!()"])', ' ', mytext))

Prints:

Did he eat it ( the bug ) ?

Upvotes: 2

brandonscript
brandonscript

Reputation: 72885

You would reference it with groups, like so (using your code as an example):

match = re.search(r'[]?!()";{}', mytext)
if match:
    replace(match.group(0), match.group(0) + " ")

You can find more information here.

Upvotes: 2

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