On 17 Sep, 19:59, Peter Otten <__pete...@web.de> wrote: > Jon Clements wrote: > > (I reckon this is probably a question for MRAB and is not really > > Python specific, but anyhow...) > > > Absolutely basic example: re.sub(r'(\d+)', r'\1', 'string1') > > > I've been searching around and I'm sure it'll be obvious when it's > > pointed out, but how do I use the above to replace 1 with 11? > > Obviously I can't use r'\11' because there is no group 11. I know I > > can use a function to do it, but it seems to me there must be a way > > without. Can I escape r'\11' somehow so that it's group 1 with a '1' > > after it (not group 11). > > Quoting > > http://docs.python.org/library/re.html#re.sub > > """ > In addition to character escapes and backreferences as described above, > \g<name> will use the substring matched by the group named name, as defined > by the (?P<name>...) syntax. \g<number> uses the corresponding group number; > \g<2> is therefore equivalent to \2, but isn’t ambiguous in a replacement > such as \g<2>0. \20 would be interpreted as a reference to group 20, not a > reference to group 2 followed by the literal character '0'. The > backreference \g<0> substitutes in the entire substring matched by the RE. > """ > > Peter
Thanks Peter and MRAB. I must have been through the docs half a dozen times and missed that - what a muppet! One of those days I guess... Cheers, Jon. -- http://mail.python.org/mailman/listinfo/python-list
