current community

your communities

more stack exchange communities

company blog
Stack Exchange Inbox Reputation and Badges
tour help
Stack Overflow is a community of 4.7 million programmers, just like you, helping each other.

Join them; it only takes a minute:

Sign up
Join the Stack Overflow community to:
  1. Ask programming questions
  2. Answer and help your peers
  3. Get recognized for your expertise
up vote 10 down vote favorite
1

What regex can I use to match ".#,#." within a string. It may or may not exist in the string. Some examples with expected outputs might be:

Test1.0,0.csv      -> ('Test1', '0,0', 'csv')         (Basic Example)
Test2.wma          -> ('Test2', 'wma')                (No Match)
Test3.1100,456.jpg -> ('Test3', '1100,456', 'jpg')    (Basic with Large Number)
T.E.S.T.4.5,6.png  -> ('T.E.S.T.4', '5,6', 'png')     (Doesn't strip all periods)
Test5,7,8.sss      -> ('Test5,7,8', 'sss')            (No Match)
Test6.2,3,4.png    -> ('Test6.2,3,4', 'png')          (No Match, to many commas)
Test7.5,6.7,8.test -> ('Test7', '5,6', '7,8', 'test') (Double Match?)

The last one isn't too important and I would only expect that .#,#. would appear once. Most files I'm processing, I would expect to fall into the first through fourth examples, so I'm most interested in those.

Thanks for the help!

share|improve this question
4  
Awww man. If only everyone would provide such an extensive list of examples that match and examples that fail... – Martin Büttner Sep 26 '12 at 18:35
    
@m.buettner I know, this is beautiful in comparison to 99% of regex questions – JKirchartz Sep 26 '12 at 18:39
up vote 3 down vote accepted

To allow for multiple consecutive matches, use lookahead/lookbehind:

r'(?<=\.)\d+,\d+(?=\.)'

Example:

>>> re.findall(r'(?<=\.)\d+,\d+(?=\.)', 'Test7.5,6.7,8.test')
['5,6', '7,8']

We can also use lookahead to perform the split as you want it:

import re
def split_it(s):
    pieces = re.split(r'\.(?=\d+,\d+\.)', s)
    pieces[-1:] = pieces[-1].rsplit('.', 1) # split off extension
    return pieces

Testing:

>>> print split_it('Test1.0,0.csv')
['Test1', '0,0', 'csv']
>>> print split_it('Test2.wma')
['Test2', 'wma']
>>> print split_it('Test3.1100,456.jpg')
['Test3', '1100,456', 'jpg']
>>> print split_it('T.E.S.T.4.5,6.png')
['T.E.S.T.4', '5,6', 'png']
>>> print split_it('Test5,7,8.sss')
['Test5,7,8', 'sss']
>>> print split_it('Test6.2,3,4.png')
['Test6.2,3,4', 'png']
>>> print split_it('Test7.5,6.7,8.test')
['Test7', '5,6', '7,8', 'test']
share|improve this answer
    
+1, this is a nice use of lookaheads. – Andrew Clark Sep 26 '12 at 19:01
    
Ya, very nice use of lookahead, thanks! – Scott B Sep 27 '12 at 19:46
up vote 4 down vote

You can use the regex \.\d+,\d+\. to find all matches for that pattern, but you will need to do a little extra to get the output you expect, especially since you want to treat .5,6.7,8. as two matches.

Here is one potential solution:

def transform(s):
    s = re.sub(r'(\.\d+,\d+)+\.', lambda m: m.group(0).replace('.', '\n'), s)
    return tuple(s.split('\n'))

Examples:

>>> transform('Test1.0,0.csv')
('Test1', '0,0', 'csv')
>>> transform('Test2.wma')
('Test2.wma',)
>>> transform('Test3.1100,456.jpg')
('Test3', '1100,456', 'jpg')
>>> transform('T.E.S.T.4.5,6.png')
('T.E.S.T.4', '5,6', 'png')
>>> transform('Test5,7,8.sss')
('Test5,7,8.sss',)
>>> transform('Test6.2,3,4.png')
('Test6.2,3,4.png',)
>>> transform('Test7.5,6.7,8.test')
('Test7', '5,6', '7,8', 'test')

To also get the file extension split off when there are no matches, you can use the following:

def transform(s):
    s = re.sub(r'(\.\d+,\d+)+\.', lambda m: m.group(0).replace('.', '\n'), s)
    groups = s.split('\n')
    groups[-1:] = groups[-1].rsplit('.', 1)
    return tuple(groups)

This will be the same output as above except that 'Test2.wma' becomes ('Test2', 'wma'), with similar behavior for 'Test5,7,8.sss' and 'Test5,7,8.sss'.

share|improve this answer
    
I was gonna suggest that... here's a link to a demo: regexr.com?329f8 – JKirchartz Sep 26 '12 at 18:46
    
Doesn't quite work if there are spaces in the name... – nneonneo Sep 26 '12 at 18:53
    
Also if the last group contains more than one . you will end up splitting the last group several times. – nneonneo Sep 26 '12 at 18:54
    
Just modified it to use \n instead of a space, you could also use something like \x00 to be more sure it won't be included in a valid string. – Andrew Clark Sep 26 '12 at 18:56
    
transform('.a.a.a.a.a.a.') == ('', 'a', 'a', 'a', 'a', 'a', 'a', '') – nneonneo Sep 26 '12 at 18:57
up vote 0 down vote 
'/^(.+)\.((\d+,\d+)\.)?(.+)$/'

The third capturing group should contain the pair of numbers. If you have multiple of those pairs, you should get multiple matches. And the third capturing would always contain the pair.

share|improve this answer
up vote 0 down vote 
^(.*?)\.(\d+,\d+)\.(.*?)$

This passes your tests, at least in Patterns:

Passing tests in Patterns

share|improve this answer
    
What's Patterns? – Burhan Khalid Sep 26 '12 at 19:26
    
itunes.apple.com/us/app/patterns-the-regex-app/id429449079 – David Eyk Sep 26 '12 at 19:56
up vote 0 down vote

This is pretty close, does python support named groups?

^.*(?P<group1>\d+(?:,\d+)?)\.(?P<group2>\d+(?:,\d+)?).*\..+$
share|improve this answer
    
The named group syntax is (?P<name>pattern) – David Eyk Sep 26 '12 at 18:48
up vote 0 down vote

Use regex pattern ^([^,]+)\.(\d+,\d+)\.([^,.]+)$

Check this demo >>

>>> print re.findall(r'^([^,]+)\.(\d+,\d+)\.([^,.]+)$', 'Test1.0,0.csv')
[('Test1', '0,0', 'csv')]

>>> print re.findall(r'^([^,]+)\.(\d+,\d+)\.([^,.]+)$', 'Test2.wma')
[]

>>> print re.findall(r'^([^,]+)\.(\d+,\d+)\.([^,.]+)$', 'Test3.1100,456.jpg')
[('Test3', '1100,456', 'jpg')]

>>> print re.findall(r'^([^,]+)\.(\d+,\d+)\.([^,.]+)$', 'T.E.S.T.4.5,6.png')
[('T.E.S.T.4', '5,6', 'png')]

>>> print re.findall(r'^([^,]+)\.(\d+,\d+)\.([^,.]+)$', 'Test5,7,8.sss')
[]

>>> print re.findall(r'^([^,]+)\.(\d+,\d+)\.([^,.]+)$', 'Test6.2,3,4.png')
[]

>>> print re.findall(r'^([^,]+)\.(\d+,\d+)\.([^,.]+)$', 'Test7.5,6.7,8.test') 
[]
share|improve this answer
    
What does this produce for: Test.xx,yz.csv? – Dave Sep 26 '12 at 18:46

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.