replacing only single instances of a character with python regexp
I am trying to replace single
$ characters with something else, and want to ignore multiple
$ characters in a row, and I can’t quite figure out how. I tried using lookahead:
s='$a $$b $$$c $d' re.sub('$(?!$)','z',s)
This gives me:
'za $zb $$zc zd'
when what I want is
'za $$b $$$c zd'
What am I doing wrong?
notes, if not using a callable for the replacement function:
- you would need look-ahead because you must not match if followed by
- you would need look-behind because you must not match if preceded by
not as elegant but this is very readable:
def dollar_repl(matchobj): val = matchobj.group(0) if val == '$': val = 'z' return val import re s = '$a $$b $$$c $d' re.sub('$+', dollar_repl, s) 'za $$b $$$c zd'
Hmm. It looks like I can get it to work if I used both lookahead and lookbehind. Seems like there should be an easier way, though.
'(?<!$)$(?!$)','z',s) 'za $$b $$$c zd're.sub(
Ok, without lookaround and without callback function:
re.sub('(^|[^$])$([^$]|$)', '1z2', s)
An alternative with
''.join('z' if x == '$' else x for x in re.split('($+)', s))