I have a regex match object in Python. I want to get the text it matched. Say if the pattern is '1.3', and the search string is 'abc123xyz', I want to get '123'. How can I do that?
I know I can use match.string[match.start():match.end()], but I find that to be quite cumbersome (and in some cases wasteful) for such a basic query.
Is there a simpler way?
You can simply use the match object's group function, like:
match = re.search(r"1.3", "abc123xyz")
if match:
doSomethingWith(match.group(0))
to get the entire match. EDIT: as thg435 points out, you can also omit the 0 and just call match.group().
Addtional note: if your pattern contains parentheses, you can even get these submatches, by passing 1, 2 and so on to group().
\1, \2 and so on (or $1, $2, depending on implementation), and the entire match is often referred to by using \0 (or $0). So it's consistent with that numbering of groups.
0, just match.group().You need to put the regex inside "()" to be able to get that part
>>> var = 'abc123xyz'
>>> exp = re.compile(".*(1.3).*")
>>> exp.match(var)
<_sre.SRE_Match object at 0x691738>
>>> exp.match(var).groups()
('123',)
>>> exp.match(var).group(0)
'abc123xyz'
>>> exp.match(var).group(1)
'123'
or else it will not return anything:
>>> var = 'abc123xyz'
>>> exp = re.compile("1.3")
>>> print exp.match(var)
None
.*(...).* for substring matches is a good practice. Just use search instead of match and it does all the for you. In fact, it's even superior, as you won't be able to get multiple matches with match (instead it will return the last one, which is also quite intuitive, as opposed to the first one).
re.finditer()MatchObjects, but not how to easily get their contents.