Using re - side effects or misunderstanding

[Sune Kirkeby]

[ "Andrew Henshaw" <andrew_dot_henshaw_at_earthling_dot_net> ]
> But I thought that ?: is for matching but not returning.  And so it is,
> under certain circumstances, e.g.

It always is, and it always does.

> if my pattern is
>     '(ab)(c)xyz'
> I get
>     [('ab', 'c')]   (Yikes! a tuple. I'm going to have to change my code a
> bit to handle this)
> but
>     '(ab)(?:c)xyz'
> yields,
>     ['ab']
> and
>     '(?:ab)(?:c)xyz'
> gives
>     ['abcxyz']

All of the above are what one would expect, since re.findall returns
a list of matches, if there are no groups in the re.  But if there are
groups it will _only_ return a list of tuples with the matched groups.

Note that (?:...) is non-grouping, so in '(?:ab)(?:c)xyz' there are
no groups, but in '(ab)(?:c)xyz' there is one group, which will then
be returned.

> so how do I get the result
>     ['abxyz']
> ??

Something along the lines of,

>>> r = re.compile('(ab)c(xyz)')
>>> matches = r.findall('..abcxyz..')
>>> matches
[('ab', 'xyz')]

almost there, just have to join the tuples,

>>> map(lambda l: string.join(l, ''), matches)
['abxyz']

Voila!

> In other words, adding groups for the purpose of adding repetitions seems to
> have a greater side-effect than I would desire.  Is there something that I'm
> missing in my use of re's?

Yes, you weren't using groups (all the time, anyway)  :-).

first-posting-ly yrs'

-- 
Sune Kirkeby                    | "Imagine, if you will, that there were no
http://mel.interspace.dk/~sune/ | such thing as a hypothetical situation..."