Python regex error: look-behind requires fixed-width pattern

Question

The following regex is supposed to match any :text: that's preceeded by start-of-string, whitespace or :, and succeeded by end-of-string, whitespace or : (Along with a few extra rules)

I'm not great at regex but I've come up with the desired solution in regexr.com:

(?<=\s|:|^)(:[^\s|:]+:)(?=\s|:|$)

:match1::match2: :match3:
:match4:
000:matchNot:
:matchNot:000
:match Not:

Result: :match1:, :match2:, :match3:, :match4:

But on Python 3 this raises an error.

re.search("(?<=\s|:|^)(:[^\s|:]+:)(?=\s|:|$)", txt)

re.error: look-behind requires fixed-width pattern

Anyone know a good workaround for this issue? Any tips are appreciated.

Make sure you use (?<![^\s:]):[^\s:]+:(?![^\s:]) and not (?:^|(?<=[\s:]))(:[^\s:]+:)(?=[\s:]|$). — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Oct 9, 2019 at 20:09
Does this answer your question? Python Regex Engine - "look-behind requires fixed-width pattern" Error — oguz ismail
– oguz ismail, Commented Feb 28, 2021 at 20:34

Jan · Accepted Answer · 2019-10-09 18:40:01Z

13

Possibly the easiest solution would be to use the newer regex module which supports infinite lookbehinds:

import regex as re

data = """:match1::match2: :match3:
:match4:
000:matchNot:
:matchNot:000
:match Not:"""

for match in re.finditer("(?<=\s|:|^)(:[^\s|:]+:)(?=\s|:|$)", data):
    print(match.group(0))

This yields

:match1:
:match2:
:match3:
:match4:

answered Oct 9, 2019 at 18:40

Jan

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3 Comments

Justas Over a year ago

This is definitely easier than splitting up look-behinds. If you can use a different library, of course.

Frankie Drake Over a year ago

This just saved me from my hell-long pattern to break :)

Rafs Over a year ago

I was using pandas str.findall and it apparently uses re under the bonnet. Had to use regex with .apply instead, but it works now.

anubhava · Accepted Answer · 2019-10-09 18:40:30Z

7

In python, you may use this work-around to avoid this error:

(?:^|(?<=[\s:]))(:[^\s:]+:)(?=[\s:]|$)

Anchors ^ and $ are zero-width matchers anyway.

RegEx Demo

answered Oct 9, 2019 at 18:40

anubhava

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Comments

Community · Accepted Answer · 2020-06-20 09:12:55Z

1

Another option would be to install regex:

$ pip3 install regex

then, we'd write some expression and (*SKIP)(*FAIL) the patterns that we wouldn't want to be there:

import regex as re

expression = r'(?:^\d+:[^:\r\n]+:$|^:[^:\r\n]+:\d+$|^(?!.*:\b\S+\b:).*$)(*SKIP)(*FAIL)|:[a-z0-9]+:'
string = '''
:match1::match2: :match3:
:match4:
000:matchNot:
:matchNot:000
:match Not:

'''

print(re.findall(expression, string))

Output

[':match1:', ':match2:', ':match3:', ':match4:']

If you wish to simplify/modify/explore the expression, it's been explained on the top right panel of regex101.com. If you'd like, you can also watch in this link, how it would match against some sample inputs.

RegEx Circuit

jex.im visualizes regular expressions:

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Oct 9, 2019 at 19:14

Emma Marcier

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