But I only want to ignore "B" if "A" is a match. If "A" is not a match,
I'd like it to advance on to "B".
On Mar 9, 2005, at 12:07 PM, Marcos Mendonça wrote:
Hi
Not and regexp expert. But it seems to me that if you want to ignora
"B" then it should be
(A) | (^B)
Hope it helps!
On Wed, 9 Mar 2005 11:11:57 -0800, Mike Hall
<[EMAIL PROTECTED]> wrote:
I'm having some strange results using the "or" operator. In every
test
I do I'm matching both sides of the "|" metacharacter, not one or the
other as all documentation says it should be (the parser supposedly
scans left to right, using the first match it finds and ignoring the
rest). It should only go beyond the "|" if there was no match found
before it, no?
Correct me if I'm wrong, but your regex is saying "match dog, unless
it's followed by cat. if it is followed by cat there is no match on
this side of the "|" at which point we advance past it and look at the
alternative expression which says to match in front of cat."
However, if I run a .sub using your regex on a string contain both dog
and cat, both will be replaced.
A simple example will show what I mean:
import re
x = re.compile(r"(A) | (B)")
s = "X R A Y B E"
r = x.sub("13", s)
print r
X R 13Y13 E
...so unless I'm understanding it wrong, "B" is supposed to be ignored
if "A" is matched, yet I get both matched. I get the same result if I
put "A" and "B" within the same group.
On Mar 8, 2005, at 6:47 PM, Danny Yoo wrote:
Regular expressions are a little evil at times; here's what I think
you're
thinking of:
###
import re
pattern = re.compile(r"""dog(?!cat)
... | (?<=dogcat)""", re.VERBOSE)
pattern.match('dogman').start()
0
pattern.search('dogcatcher').start()
Hi Mike,
Gaaah, bad copy-and-paste. The example with 'dogcatcher' actually
does
come up with a result:
###
pattern.search('dogcatcher').start()
6
###
Sorry about that!
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