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(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 -- http://mail.python.org/mailman/listinfo/python-list
