Python Regex - How to Get Positions and Values of Matches

How can I get the start and end positions of all matches using the re module? For example given the pattern r'[a-z]' and the string 'a1b2c3d4' I'd want to get the positions where it finds each letter. Ideally, I'd like to get the text of the match back too.

import re
p = re.compile("[a-z]")
for m in p.finditer('a1b2c3d4'):
    print(m.start(), m.group())

Taken from

Regular Expression HOWTO

> span() returns both start and end indexes in a single tuple. Since the > match method only checks if the RE matches at the start of a string, > start() will always be zero. However, the search method of RegexObject > instances scans through the string, so the match may not start at zero > in that case. >

>>> p = re.compile('[a-z]+')
>>> print p.match('::: message')
None
>>> m = p.search('::: message') ; print m
<re.MatchObject instance at 80c9650>
>>> m.group()
'message'
>>> m.span()
(4, 11)

Combine that with:

In Python 2.2, the finditer() method is also available, returning a sequence of MatchObject instances as an iterator.

>>> p = re.compile( ... )
>>> iterator = p.finditer('12 drummers drumming, 11 ... 10 ...')
>>> iterator
<callable-iterator object at 0x401833ac>
>>> for match in iterator:
...     print match.span()
...
(0, 2)
(22, 24)
(29, 31)

you should be able to do something on the order of

for match in re.finditer(r'[a-z]', 'a1b2c3d4'):
   print match.span()

For Python 3.x

from re import finditer
for match in finditer("pattern", "string"):
    print(match.span(), match.group())

You shall get \n separated tuples (comprising first and last indices of the match, respectively) and the match itself, for each hit in the string.

note that the span & group are indexed for multi capture groups in a regex

regex_with_3_groups=r"([a-z])([0-9]+)([A-Z])"
for match in re.finditer(regex_with_3_groups, string):
    for idx in range(0, 4):
        print(match.span(idx), match.group(idx))