Nested regular expression strings

Question

I'm having trouble writing a regular expression for the following. I have a vector of literals (see RE_LIT) and I would like to find all the vectors in a line of text but I'm having difficulty writing the regular expression. Specifically I seem to have issues with the parenthesis acting as groups and not parenthesis.

RE_LABEL1 = r'[cvx]\d+(?![.]r)$'
RE_LABEL2 = r'v\d+\.r\d+'
RE_LABEL = r'(%s)|(%s)' % (RE_LABEL1, RE_LABEL2)
RE_LIT = r'!?%s' % RE_LABEL
RE_VEC = r'\[\s*(\s*%s\s*,?\s*)+\s*\]' % RE_LIT

Example string to match:

test = 'c1 = blah([v3,v4,v5.r1,!v6,v7,x8,v9,v10], [v1, v2], [x5.r1])'

Expected results:

> print re.findall(RE_VEC, test)
['[v3,v4,v5.r1,!v6,v7,x8,v9,v10]', '[v1, v2]']

Thank you ahead of time for your help.

"I seem to have issues with the parenthesis acting as groups and not parenthesis" - parentheses have meaning in regular expressions. If you want them to just be characters, escape them with a prefixed backslash: \(. Your example isn't a very useful test, as r'\[[^]]+]' would work fine... — jonrsharpe
– jonrsharpe, Commented Apr 28, 2015 at 21:26
Jonrsharpe: If I were to use this method how would I ensure that the entry's in the list are literals (RE_LIT) not just characters? — fcimeson
– fcimeson, Commented Apr 28, 2015 at 22:40
Yes parentheses have meaning in regular expressions but is there some mechanism that will act like parentheses normally does for nesting? — fcimeson
– fcimeson, Commented Apr 28, 2015 at 22:43

Wiktor Stribiżew · Accepted Answer · 2015-04-28 21:44:03Z

1

You can use the following fix:

import re
RE_LABEL1 = r'[cvx]\d+(?![.]r)'
RE_LABEL2 = r'v\d+\.r\d+'
RE_LABEL = r'%s|%s' % (RE_LABEL1, RE_LABEL2)
RE_LIT = r'\!?%s),?\s*' % RE_LABEL
RE_VEC = r'(?:(?:%s)+' % RE_LIT
test = '[v3,v4,v5.r1,!v6,v7,x8,v9,v10], [v1, v2]'
print re.findall(RE_VEC, test)

Output of an IDEONE demo:

['v3,v4,v5.r1,!v6,v7,x8,v9,v10', 'v1, v2']

answered Apr 28, 2015 at 21:44

Wiktor Stribiżew

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Wiktor Stribiżew Over a year ago

Does this work for you or do you need another approach?

fcimeson · Accepted Answer · 2015-04-29 11:59:46Z

0

import re
RE_LABEL1 = r'[cvx]\d+(?=[ ,\]])'
RE_LABEL2 = r'v\d+\.r\d+(?=[ ,\]])'
RE_LABEL = r'%s|%s' % (RE_LABEL1, RE_LABEL2)
RE_LIT = r'\!?%s' % RE_LABEL
RE_VEC = r'\[\s*(?:(?:\s*%s\s*\s*),?)+\s*\]' % RE_LIT
test = '[v3,v4,v5.r1,!v6,v7,x8,v9,v10], [v1, v2], [v1, x2.r2]'
print re.findall(RE_VEC, test)

Thank you stribizhev for your help, it got me half of the way there. The above is the finial solution.

answered Apr 29, 2015 at 11:59

fcimeson

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