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Well its this normal? i want to concatenate a number to a
backreference in a regular expression. Im working in a multprocess
script so the first what i think is in an error in the multiprocess
logic but what a sorprise!!! when arrived to this conclussion after
some time debugging i see that:

import re
aa = "zzzxx"
re.sub(r'(zzz.*',r'\1'+str(3333),aa)
'[33'

¿?¿?¿? well lets put a : after the backreference

aa = "zzzxx"
re.sub(r'(zzz).*',r'\1:'+str(3333),aa)
'zzz:3333'

now its the expected result.... so
should i expect that python concatenate the string to the
backreference before substitute the backreference? or is a bug

tested on:
Python 2.6.2 (r262:71605, Apr 14 2009, 22:40:02) [MSC v.1500 32 bit
(Intel)] on win32
Python 2.5.2 (r252:60911, Jan 4 2009, 17:40:26) [GCC 4.3.2] on linux2

with the same result

Cheers!

abdulet wrote:

> Well its this normal? i want to concatenate a number to a
> backreference in a regular expression. Im working in a multprocess
> script so the first what i think is in an error in the multiprocess
> logic but what a sorprise!!! when arrived to this conclussion after
> some time debugging i see that:
>
> import re
> aa = "zzzxx"
> re.sub(r'(zzz.*',r'\1'+str(3333),aa)
> '[33'

If you perform the addition you get r"\13333". How should the regular
expression engine interpret that? As the backreference to group 1, 13, ...
or 13333? It picks something completely different, "[33", because "\133" is
the octal escape sequence for "[":

>>> chr(0133)
'['

You can avoid the ambiguity with

extra = str(number)
extra = re.escape(extra)
re.sub(expr r"\g<1>" + extra, text)

The re.escape() step is not necessary here, but a good idea in the general
case when extra is an arbitrary string.

Peter

On 23 oct, 13:54, Peter Otten <__pete...@web.de> wrote:
> abdulet wrote:
> > Well its this normal? i want to concatenate a number to a
> > backreference in a regular expression. Im working in a multprocess
> > script so the first what i think is in an error in the multiprocess
> > logic but what a sorprise!!! when arrived to this conclussion after
> > some time debugging i see that:
>
> > import re
> > aa = "zzzxx"
> > re.sub(r'(zzz.*',r'\1'+str(3333),aa)
> > '[33'
>
> If you perform the addition you get r"\13333". How should the regular
> expression engine interpret that? As the backreference to group 1, 13, ...
> or 13333? It picks something completely different, "[33", because "\133" is
> the octal escape sequence for "[":
>
> >>> chr(0133)
>
> '['
>
> You can avoid the ambiguity with
>
> extra = str(number)
> extra = re.escape(extra)
> re.sub(expr r"\g<1>" + extra, text)
>
> The re.escape() step is not necessary here, but a good idea in the general
> case when extra is an arbitrary string.
>
> Peter
Aha!!! nice thanks i don't see that part of the re module
documentation and it was in front of my eyes ( like always its
something silly jjj so thanks again and yes!! is a nice idea to escape
the variable

cheers