Python regex non greedy

Question

as we know the python '?' quantifier is consider a lazy quantifier and then should gives us the shortest match, but in my example i am getting the second match(second_occurrence), instead of the first match(first_occurrence)

content = "this is how we want that first_occurrence over there but that second_occurrence it is 
always wrong when "
match = re.search(r"^this .* that (?P<occurrence>.*?) ", content)
print(match.groupdict())

>>> match = re.search(r"^(this .* that )(?P<occurrence>.*?) ", content) >>> print(match.group(1)) this is how we want that first_occurrence over there but that — user12097764
– user12097764, Commented Dec 4, 2019 at 18:58

Andrej Kesely · Accepted Answer · 2019-12-04 18:41:28Z

1

In your expression "^this .* that (?P<occurrence>.*?) " the first .* is greedy, so that it will match all the way to the last that.

Change your example to:

import re

content = "this is how we want that first_occurrence over there but that second_occurrence it is always wrong when "
match = re.search(r"^this .*? that (?P<occurrence>.*?) ", content)
print(match.groupdict())

This prints:

{'occurrence': 'first_occurrence'}

answered Dec 4, 2019 at 18:41

Andrej Kesely

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