[python] 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.

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.

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.

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()

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))

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()

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))

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()