Python Regex get everything within parentheses unless in quotes

Question

Given the string

S = "(45171924,-1,'AbuseFilter/658',2600),(43795362,-1,'!!_(disambiguation)',2600),(45795362,-1,'!!_(disambiguation)',2699)"

I'd like to extract everything within the parentheses UNLESS the parens are inside a quotation. So far I've managed to get everything within parentheses, but I can't figure out how to stop from splitting on the inner parenthesis inside the quotes. My current code is:

import re
S = "(45171924,-1,'AbuseFilter/658',2600),(43795362,-1,'!!_(disambiguation)',2600),(45795362,-1,'!!_(disambiguation)',2699)"

p = re.compile( "\((.*?)\)" )
m =p.findall(S)
for element in m:
    print element

What I want is:

45171924,-1,'AbuseFilter/658',2600
43795362,-1,'!!_(disambiguation)',2600
45795362,-1,'!!_(disambiguation)',2699

What I currently get is:

45171924,-1,'AbuseFilter/658',2600
43795362,-1,'!!_(disambiguation
45795362,-1,'!!_(disambiguation

What can I do in order to ignore the internal paren?

Thank you!!

In case it helps, here are the threads I've looked at:

1) REGEX-String and escaped quote

2) Regular expression to return text between parenthesis

3)Get the string within brackets in Python

hwnd · Answer 1 · 2015-04-19T04:41:41.713

3

You can use a non-capturing group to assert either a comma or the end of the string follows:

p = re.compile(r'\((.*?)\)(?:,|$)')

Working Demo

edited Apr 19 '15 at 04:41

answered Apr 19 '15 at 04:33

hwnd

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score 1 · Answer 2 · answered Apr 19 '15 at 04:46

1

for element in S[1:-1].split('),('):
    print element

answered Apr 19 '15 at 04:46

mmachine

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Avinash Raj · Answer 3 · 2015-04-19T06:44:12.673

You could use the below regex.

>>> import re
>>> s = "(45171924,-1,'AbuseFilter/658',2600),(43795362,-1,'!!_(disambiguation)',2600),(45795362,-1,'!!_(disambiguation)',2699)"
>>> for i in re.findall(r"\(((?:'[^']*'|[^()])*)\)", s):
        print(i)


45171924,-1,'AbuseFilter/658',2600
43795362,-1,'!!_(disambiguation)',2600
45795362,-1,'!!_(disambiguation)',2699

Explanation:

\( - Matches a literal ( symbol.
( - Start of a capturing group.
(?:'[^']*'|[^()])* - '[^']*' part matches greedily the single quoted block. If there is any (, ) symbols present inside that, it won't care about that. Because we used [^']* which matches any character but not of ' , zero or more times. If the following character is not the start of a single quoted block then the control transfers to the pattern which exists next to the | symbol ie, [^()] which matches any character but not of ( or ). So the whole (?:'[^']*'|[^()])* will match a single quoted block or any char not of (, ) , zero or more times.
) end of the capturing group.
\) literal ) symbol.

DEMO

Thank you! This was super helpful and general enough all different string permutations. — MayaR, Apr 22 '15 at 20:26
@MayaR [Accept an answer which you loves the most](http://stackoverflow.com/help/accepted-answer). — Avinash Raj, Apr 23 '15 at 00:42

score 0 · Answer 4 · answered Apr 19 '15 at 04:40

Some simple approach would be negative lookahead - check that after closing brace no quote follows, e.g.

import re
S = "(45171924,-1,'AbuseFilter/658',2600),(43795362,-1,'!!_(disambiguation)',2600),(45795362,-1,'!!_(disambiguation)',2699)"

m = re.findall(r'\((.*?)\)(?![\'])', S)
for element in m:
    print element

prints

45171924,-1,'AbuseFilter/658',2600
43795362,-1,'!!_(disambiguation)',2600
45795362,-1,'!!_(disambiguation)',2699

http://www.codeskulptor.org/#user39_CL89xhroV0_0.py

I have put the quote in character class (square brackets), so that you could add other symbols, which should make the closing bracket being ignored.

Python Regex get everything within parentheses unless in quotes

4 Answers4