How to reference a regex match as a variable in python

Question

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.

Community · Accepted Answer · 2017-05-23 12:13:06Z

3

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.}

edited May 23, 2017 at 12:13

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answered Dec 3, 2013 at 5:17

p.s.w.g

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brandonscript · Accepted Answer · 2013-12-03 05:13:31Z

2

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.

answered Dec 3, 2013 at 5:13

brandonscript

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perreal · Accepted Answer · 2013-12-03 05:28:49Z

2

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 ) ?

edited Dec 3, 2013 at 5:28

answered Dec 3, 2013 at 5:22

perreal

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1 Comment

p.s.w.g Over a year ago

+1 I was trying to figure out how to do make sure you don't get multiple spaces when your answer popped up. For reason this didn't occur to me, guess I need to get some sleep :)

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