On 5/5/05, Jeremy Bowers <[EMAIL PROTECTED]> wrote:
> On Wed, 04 May 2005 20:24:51 +0800, could ildg wrote:
>
> > Thank you.
> >
> > I just learned how to use re, so I want to find a way to settle it by
> > using re. I know that split it into pieces will do it quickly.
>
> I'll say this; you have two problems, splitting out the numbers and
> verifying their conformance to some validity rule.
>
> I strongly recommend treating those two problems separately. While I'm not
> willing to guarantee that an RE can't be written for something like ("[A
> number A]_[A number B]" such that A < B) in the general case, it won't be
> anywhere near as clean or as easy to follow if you just write an RE to
> extract the numbers, then verify the constraints in conventional Python.
>
> In that case, if you know in advance that the numbers are guaranteed to be
> in that format, I'd just use the regular expression "\d+", and the
> "findall" method of the compile expression:
>
> Python 2.3.5 (#1, Mar 3 2005, 17:32:12)
> [GCC 3.4.3 (Gentoo Linux 3.4.3, ssp-3.4.3-0, pie-8.7.6.6)] on linux2
> Type "help", "copyright", "credits" or "license" for more information.
> >>> import re
> >>> m = re.compile("\d+")
> >>> m.findall("344mmm555m1111")
> ['344', '555', '1111']
> >>>
>
> If you're checking general matching of the parameters you've given, I'd
> feel no shame in checking the string against r"^(_\d+){1,3}$" with .match
> and then using the above to get the numbers, if you prefer that. (Note
> that I believe .match implies the initial ^, but I tend to write it
> anyways as a good habit. Explicit better than implicit and all that.)
>
> (I just tried to capture the three numbers by adding a parentheses set
> around the \d+ but it only gives me the first. I've never tried that
> before; is there a way to get it to give me all of them? I don't think so,
> so two REs may be required after all.)
You can capture each number by using group, each group can have a name.
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