CODEWITHSUNDEEP
regexpython-3.x
freeaks
freeaks

Reputation: 487

python regex return result without characters used for matching

i would like to perform regex matching, but exclude from the result the characters used for the matching. consider this sample code

import re
string="15  83(B): D17"
result=re.findall(r'\(.+\)',string)
print(result)

here's what i get: ['(B)']

here's what i'd like to have: ['B']

i'm after a generic solution to exclude pattern characters used to start/end a match from the result, not just a solution for this precise case. for example instead of just ( and ) i could have used more complex patterns to start/end matching, and i still would not want to see them as part of the result.

Upvotes: 0

Views: 39

Answers (3)

glibdud
glibdud

Reputation: 7840

Lookaheads and lookbehinds can be used to assert that characters are present before and after a certain position without including them in a match.

>>> string = "15  83(B): D17"
>>> re.findall(r'(?<=\().+(?=\))', string)
['B']

Here, (?<=\() is a positive lookbehind that asserts that an open parenthesis character comes immediately before this position. (?=\)) is a positive lookahead that asserts that a close parenthesis character comes immediately after.

Search the re module documentation for the terms "lookahead" and "lookbehind" for more information.

Upvotes: 0

Eduardo Lynch Araya
Eduardo Lynch Araya

Reputation: 824

The correct regex is it

r"(?<=\()(.*)(?=\))"

U can change ( and ) for anything

Upvotes: 0

anubhava
anubhava

Reputation: 784998

You need to use a capturing group for the text you need in output like this:

>>> string="15  83(B): D17"
>>> print re.findall(r'\((.*?)\)', string)
['B']

(.*?) is capturing group to match and capture 0 or more characters, non-greedy

In general you can replace starting ( and ending ) with anything you have as before and after your match.

Upvotes: 2

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