I use Python's re module to replace a substring, such as:
>>> import re
>>> re.sub(r"a.*b","ab","acbacbacb")
'ab'
This matches .* with cbacbac, but I want it to match c three times, so that the output is ababab.
Could anybody tell me how to do it?
The easiest solution would be to use the lazy (non-greedy) *? operator:
>>> re.sub(r"a.*?b","ab","acbacbacb")
'ababab'
This might however have an impact on performance. Because of the structure of this regex, you can just as well use the equivalent
re.sub(r"a[^b]*b","ab","acbacbacb")
which could perform better, depending on how good the optimizer is.
If you have even more a priori knowledge about the structure of the .* part, you should make it even more explicit. Say, for example, that you already know that between the a and the b there will be only c's, you can do
re.sub(r"ac*b","ab","acbacbacb")
Regex by default is greedy. Use .*?
>>> import re
>>> re.sub(r"a.*?b","ab","acbacbacb")
'ababab'
>>>
http://docs.python.org/library/re.html
The
*,+, and?qualifiers are all greedy; they match as much text as possible. Sometimes this behaviour isn’t desired; if the RE<.*>is matched against'<H1>title</H1>', it will match the entire string, and not just'<H1>'. Adding?after the qualifier makes it perform the match in non-greedy or minimal fashion; as few characters as possible will be matched. Using.*?in the previous expression will match only'<H1>'.
use non-greedy match:
re.sub(r"a.*?b","ab","acbacbacb")
'ababab'
from http://docs.python.org/library/re.html:
The '*', '+', and '?' qualifiers are all greedy; they match as much text as possible. Sometimes this behaviour isn’t desired. [...] Adding '?' after the qualifier makes it perform the match in non-greedy or minimal fashion; as few characters as possible will be matched.
re.sub(r"a.b","ab","acbacbacb")
'ababab'
