Non greedy regex

[tibetan.houdini]

Can someone please explain why these expressions both produce the same
result? Surely this means that non-greedy regex does not work?

print re.sub( 'a.*b', '', 'ababc' )

gives: 'c'

Understandable. But

print re.sub( 'a.*?b', '', 'ababc' )

gives: 'c'

NOT, understandable. Surely the answer should be: 'abc'

 

[Carsten Haese]

Can someone please explain why these expressions both produce the same
result? Surely this means that non-greedy regex does not work?

print re.sub( 'a.*b', '', 'ababc' )

gives: 'c'

Understandable. But

print re.sub( 'a.*?b', '', 'ababc' )

gives: 'c'

NOT, understandable. Surely the answer should be: 'abc'

You didn't tell re.sub to only substitute the first occurrence of the
pattern. It non-greedily matches two occurrences of 'ab' and replaces
each of them with ''. Try replacing with some non-empty string instead
to observe the difference between the two behaviors.

-Carsten

 

[Thomas Nelson]

There's an optional count argument that will give what you want. Try
re.sub('a.*b','','ababc',count=1)

 

[Maksim Kasimov]

> print re.sub( 'a.*?b', '', 'ababc' )
>
> gives: 'c'

you've forgot "^" at the begging of the pattern (there is another 'ab' before 'c')


print re.sub( '^a.*?b', '', 'ababc' )