[Python] Q: How to get positions of multiple RE matches?

better make that loop:

while 1:
m = p.search(a, pos)
if not m:
break
pos = m.start()
print pos
pos += 1

Cheers /F

"Karl Schmid" <schmid at ice.mpg.de> wrote in message
news:9akdum$2ve8$1 at gwdu67.gwdg.de...
Depends on your definition of 'elegance', I guess, but...:

class Matches:
def __init__(self):
self.matches = []
def __call__(self, mo):
self.matches.append(mo.start())
return ''

import re
a = '''This module provides regular expression matching
operations similar to those found in Perl'''

p = re.compile('(o)')
matches = Matches()
junk = p.sub(matches, a)

for match in matches.matches:
print match



Some would call it "elegant", some would call it "tricky"
(because the .sub is being used only for its side effects,
and its result is uninteresting and gets discarded).


Anyway, the .sub method is the only documented way I know
to get a *call-back* for every non-overlapping match, "at a
single stroke", without explicitly programming a loop
yourself. (The "non-overlapping" feature may not be
what you're looking for, of course, in which case this
approach is certainly ungood).


If I did decide I had better program a loop (e.g., I also
want matches that would overlap with others), and I'm not
being paranoic about performance (if I needed to be, I
would carefully time each possible alternative), I think
I'd code it the simplest way I can think of...:

for i in range(len(a)):
if p.match(a[i:]):
print i


Alex