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Re: Help with regular expression in python

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Hi Josh,
thanks for the reply. I am no expert so please bear with me:
I thought that the {32} was supposed to match the previous expression 32 
times?

So how can i have all matches accessible to me?

matt



On Thursday, August 18, 2011, Josh Benner wrote:
> On Thu, Aug 18, 2011 at 4:03 PM, Matt Funk <matze999@gmail.com> wrote:
> > Hi guys,
> > 
> > thanks for the suggestions. I had tried the white space before as well
> > (to no
> > avail). So here is the expression i am using (based on suggestions), but
> > still
> > no success:
> > 
> > instance_linetype_pattern_str =\
> > 
> >        r'(([-+]?(\d+(\.\d*)?|\.\d+)([eE][-+]?\d+))?\s+){32}(.+)'
> > 
> > instance_linetype_pattern = re.compile(instance_linetype_pattern_str)
> > results = instance_linetype_pattern.findall(line)
> > print "results: "; print results
> > 
> > 
> > The match i get is:
> > results:
> > [('2.199000e+01 ', '2.199000', '.199000', 'e+01', ': (instance:
> > 0)\t:\tsome description')]
> > 
> > 
> > btw: The line to be matched (given below) is ONE line. There are no line
> > breaks (even though my email client adds them).
> > 
> > 
> > matt
> 
> If a group matches multiple times, only the last match is accessible.  The
> matches returned represent the inner groupings of the last match found.
> 
> JB-)

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Re: Help with regular expression in python Matt Funk <matze999@gmail.com> - 2011-08-19 08:09 -0600

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