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Groups > comp.lang.python > #97255

Re: Question about regular expression

Date 2015-09-30 14:20 -0500
From Tim Chase <python.list@tim.thechases.com>
Subject Re: Question about regular expression
References <811788b6-9955-4dcc-bf49-9647891d17ec@googlegroups.com>
Newsgroups comp.lang.python
Message-ID <mailman.276.1443641310.28679.python-list@python.org> (permalink)

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On 2015-09-30 11:34, massi_srb@msn.com wrote:
> firstly the description of my problem. I have a string in the
> following form:
> 
> s = "name1 name2(1) name3 name4 (1, 4) name5(2) ..."
> 
> that is a string made up of groups in the form 'name' (letters
> only) plus possibly a tuple containing 1 or 2 integer values.
> Blanks can be placed between names and tuples or not, but they
> surely are placed beween two groups. I would like to process this
> string in order to get a dictionary like this:
> 
> d = {
>     "name1":(0, 0),
>     "name2":(1, 0),
>     "name3":(0, 0),
>     "name4":(1, 4),
>     "name5":(2, 0),
> }
> 
> I guess this problem can be tackled with regular expressions, b

First out of the gate, I suggest you follow Emile's advice and try
using string expressions.  However, if you *want* to do it with
regular expressions, you can.  It's ugly and might be fragile, but

#############################################################
import re
s = "name1 name2(1) name3 name4 (1, 4) name5(2) ..."
r = re.compile(r"""
    \b       # start at a word boundary
    (\w+)    # capture the word
    \s*      # optional whitespace
    (?:      # start an optional grouping for things in the parens
     \(      # a literal open-paren
      \s*    # optional whitespace
      (\d+)  # capture the number in those parens
      (?:    # start a second optional grouping for the stuff after a comma
       \s*   # optional whitespace
       ,     # a literal comma
       \s*   # optional whitespace
       (\d+) # the second number
      )?     # make the command and following number optional
     \)      # a literal close-paren
    )?       # make that stuff in parens optional
    """, re.X)
d = {}
for m in r.finditer(s):
    a, b, c  = m.groups()
    d[a] = (int(b or 0), int(c or 0))

from pprint import pprint
pprint(d)
#############################################################


I'd stick with the commented version of the regexp if you were to use
this anywhere so that others can follow what you're doing.

-tkc




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Thread

Question about regular expression massi_srb@msn.com - 2015-09-30 11:34 -0700
  Re: Question about regular expression Emile van Sebille <emile@fenx.com> - 2015-09-30 11:50 -0700
  Re: Question about regular expression Tim Chase <python.list@tim.thechases.com> - 2015-09-30 14:20 -0500
  Re: Question about regular expression Denis McMahon <denismfmcmahon@gmail.com> - 2015-09-30 23:30 +0000
    Re: Question about regular expression Denis McMahon <denismfmcmahon@gmail.com> - 2015-10-02 18:25 +0000
  Re: Question about regular expression Emile van Sebille <emile@fenx.com> - 2015-09-30 20:58 -0700
  Re: Question about regular expression Tim Chase <python.list@tim.thechases.com> - 2015-10-01 07:39 -0500
  Re: Question about regular expression Rob Gaddi <rgaddi@technologyhighland.invalid> - 2015-10-01 15:53 +0000
    Re: Question about regular expression Denis McMahon <denismfmcmahon@gmail.com> - 2015-10-01 21:41 +0000
  Re: Question about regular expression Denis McMahon <denismfmcmahon@gmail.com> - 2015-10-01 21:31 +0000

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