[Tutor] regexp: a bit lost

[Alex Hall]

On 10/1/10, Steven D'Aprano <steve at pearwood.info> wrote:
> On Fri, 1 Oct 2010 12:45:38 pm Alex Hall wrote:
>> Hi, once again...
>> I have a regexp that I am trying to use to make sure a line matches
>> the format: [c*]n [c*]n n
>> where c* is (optionally) 0 or more non-numeric characters and n is
>> any numeric character. The spacing should not matter. These should
>> pass: v1 v2   5
>> 2 someword7 3
>>
>> while these should not:
>> word 2  3
>> 1 2
>>
>> Here is my test:
>> s=re.search(r"[\d+\s+\d+\s+\d]", l)
>
> Try this instead:
>
> re.search(r'\d+\s+\D*\d+\s+\d', l)
>
> This searches for:
>     one or more digits
>     at least one whitespace char (space, tab, etc)
>     zero or more non-digits
>     at least one digit
>     at least one whitespace
>     exactly one digit
Makes sense.
>
>
>> However:
>> 1. this seems to pass with *any* string, even when l is a single
>> character. This causes many problems
> [...]
>
> I'm sure it does.
>
> You don't have to convince us that if the regular expression is broken,
> the rest of your code has a problem. That's a given. It's enough to
> know that the regex doesn't do what you need it to do.
Understood. My intent was to ask why my regexp would match anything at all.
>
>
>> 3. Once I get the above working, I will need a way of pulling the
>> characters out of the string and sticking them somewhere. For
>> example, if the string were
>> v9 v10 15
>> I would want an array:
>> n=[9, 10, 15]
>
>
> Modify the regex to be this:
>
> r'(\d+)\s+\D*(\d+)\s+(\d)'
>
> and then query the groups of the match object that is returned:
>
>>>> mo = re.search(r'(\d+)\s+\D*(\d+)\s+(\d)', 'spam42   eggs23    9')
>>>> mo.groups()
> ('42', '23', '9')
>
> Don't forget that mo will be None if the regex doesn't match, and don't
> forget that the items returned are strings.
Alright that worked perfectly, after a lot of calls to int()! I also
finally understand what a group is and, at a basic level, how to use
it. I have wondered how to extract matched text from a string for a
long time, and this has finally answered that. Thanks!
>
>
>
> --
> Steven D'Aprano
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-- 
Have a great day,
Alex (msg sent from GMail website)
mehgcap at gmail.com; http://www.facebook.com/mehgcap