> 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' )
[EMAIL PROTECTED] wrote:
> Can someone please explain why these expressions both produce the same
> result? Surel
There's an optional count argument that will give what you want. Try
re.sub('a.*b','','ababc',count=1)
Carsten Haese wrote:
> On Thu, 2006-12-14 at 06:45 -0800, [EMAIL PROTECTED] wrote:
> > Can someone please explain why these expressions both produce the same
> > result? Surely this means that
On Thu, 2006-12-14 at 06:45 -0800, [EMAIL PROTECTED] wrote:
> 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(
