Bug? concatenate a number to a backreference: re.sub(r'(zzz:)xxx', r'\1'+str(4444), somevar)
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 = zzz:xxx re.sub(r'(zzz:).*',r'\1'+str(),aa) '[33' ¿?¿?¿? well lets put a : after the backreference aa = zzz:xxx re.sub(r'(zzz).*',r'\1:'+str(),aa) 'zzz:' 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! -- http://mail.python.org/mailman/listinfo/python-list
Re: Bug? concatenate a number to a backreference: re.sub(r'(zzz:)xxx', r'\1'+str(4444), somevar)
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 = zzz:xxx re.sub(r'(zzz:).*',r'\1'+str(),aa) '[33' If you perform the addition you get r\1. How should the regular expression engine interpret that? As the backreference to group 1, 13, ... or 1? 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\g1 + 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 -- http://mail.python.org/mailman/listinfo/python-list
Re: Bug? concatenate a number to a backreference: re.sub(r'(zzz:)xxx', r'\1'+str(4444), somevar)
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 = zzz:xxx re.sub(r'(zzz:).*',r'\1'+str(),aa) '[33' If you perform the addition you get r\1. How should the regular expression engine interpret that? As the backreference to group 1, 13, ... or 1? 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\g1 + 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 -- http://mail.python.org/mailman/listinfo/python-list
