Help with regular expression in python

Matt Funk matze999 at gmail.com
Fri Aug 19 23:55:53 CEST 2011


On Friday, August 19, 2011, Carl Banks wrote:
> On Friday, August 19, 2011 10:33:49 AM UTC-7, Matt Funk wrote:
> > number = r"\d\.\d+e\+\d+"
> > numbersequence = r"%s( %s){31}(.+)" % (number,number)
> > instance_linetype_pattern = re.compile(numbersequence)
> > 
> > The results obtained are:
> > results:
> > [(' 2.199000e+01', ' : (instance: 0)\t:\tsome description')]
> > so this matches the last number plus the string at the end of the line,
> > but no retaining the previous numbers.
> > 
> > Anyway, i think at this point i will go another route. Not sure where the
> > issues lies at this point.
> 
> I think the problem is that repeat counts don't actually repeat the
> groupings; they just repeat the matchings.  Take this expression:
> 
> r"(\w+\s*){2}"
I see

> 
> This will match exactly two words separated by whitespace.  But the match
> result won't contain two groups; it'll only contain one group, and the
> value of that group will match only the very last thing repeated:
> 
> Python 2.7.1+ (r271:86832, Apr 11 2011, 18:13:53)
> [GCC 4.5.2] on linux2
> Type "help", "copyright", "credits" or "license" for more information.
> 
> >>> import re
> >>> m = re.match(r"(\w+\s*){2}","abc def")
> >>> m.group(1)
> 
> 'def'
> 
> So you see, the regular expression is doing what you think it is, but the
> way it forms groups is not.
> 
> 
> Just a little advice (I know you've found a different method, and that's
> good, this is for the general reader).
> 
> The functions re.findall and re.finditer could have helped here, they find
> all the matches in a string and let you iterate through them.  (findall
> returns the strings matched, and finditer returns the sequence of match
> objects.)  You could have done something like this:
I did use findall but when i tried to match the everything (including the 'some 
description' part) it did not work. But i think the explanation you gave above 
matches this case and explains why it did not.


> 
> row = [ float(x) for x in re.findall(r'\d+\.\d+e\+d+',line) ]
> 
> And regexp matching is often overkill for a particular problem; this may be
> of them.  line.split() could have been sufficient:
> 
> row = [ float(x) for x in line.split() ]
> 
> Of course, these solutions don't account for the case where you have lines,
> some of which aren't 32 floating-point numbers.  You need extra error
> handling for that, but you get the idea.

thanks
matt

> 
> 
> Carl Banks




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