Regexp question

Mitja nun at example.com
Wed Dec 1 17:43:28 CET 2004


On Wed, 1 Dec 2004 07:48:24 -0600, Philippe C. Martin  
<philippecmartin at sbcglobal.net> wrote:

> I realize this is more a regexp question than a python question, but  
> maybe one
> of the re object could help me:
>
> I have wish to know how to _no_ match:
>
> This is but an example of the data I handle:
>
> xx xx xx xx xx xx xx [yy yy yy yy yy yy yy] (zz zz zz zz)
>
> I currently can retrieve the three group of logical data blocks with:
>
> l_str = 'xx xx xx xx xx xx xx [yy yy yy yy yy yy yy] (zz zz zz zz)'
> p = re.compile(r'([a-f-0-9\s]*) (\[[a-f-0-9\s]*\])
> (\([a-f-0-9\s]*\))',re.IGNORECASE) #OK
> g = p.search(l_str)
>
>
> What I would rather do is.
>
> "get the data block that is _not_ between brackets or parenthesis  i.e;  
> 'xx xx
> xx xx xx xx xx' knowing that the intial string could be:
>
> [yy yy yy yy yy yy yy]  xx xx xx xx xx xx xx  (zz zz zz zz)
>
>
> Any clue  ?

regexps seem an overkill for the task at hand.

If data is really as simple as you suggest, you can try the following:
>>> s = 'xx [y y] (z z)'
>>> s = s[:s.index('(')] + s[s.index(')')+1:]
>>> s
'xx [y y] '
>>> s = s[:s.index('[')] + s[s.index(']')+1:]
>>> s
'xx  '
>>> s.strip()
'xx'


Relevant lines:
s = s[:s.index('(')] + s[s.index(')'):]
s = s[:s.index('[')] + s[s.index(']')+1:]
s = s.strip()
-- 
Mitja



More information about the Python-list mailing list