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( '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 -- http://mail.python.org/mailman/listinfo/python-list
