regex result filter in python

Question

I am working on a program with regexes, I have to filter them but I can't find out how. I want to match every red,xxxx or xxxx,red expression in my string and put the colors xxxx into a group. Here is my code:

string = "blue,red   red,yellow   blue,yellow   red,green   purple red, ..."
regex = re.compile('(?:red,(?P<redfirst>\w+)|(?P<othercolorfirst>\w+),red)')

Then I write:

for match in regex.finditer(string):
    if match.group('redfirst')!= "None":
        print(match.group("redfirst"))

But I still obtain printing like:

None
yellow
green
None

I dont want the 'None' results to appear, I have to skip them in an smart way if possible. Thanks for help!

EDIT None without quotes doesn't work either

Ben · Accepted Answer · 2011-08-30 19:31:55Z

3

>>> import re
>>> regex = re.compile('(?:red,(?P<redfirst>\w+)|(?P<othercolorfirst>\w+),red)')

>>> string = "blue,red   red,yellow   blue,yellow   red,green   purple red, ..."

>>> for matches in regex.finditer(string):
...     if matches.group('redfirst'):
...         print matches.group('redfirst')
...
yellow
green
>>>

answered Aug 30, 2011 at 19:31

Ben

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user395760 · Accepted Answer · 2011-08-30 19:30:44Z

2

The result when nothing matches isn't "None" (the string), it's None (the singleton object). And while simply stripping the quotes around None in your condition works, it's preferred to use ... is None for numerous reasons, the most important being that it's in the style guide (hey, consistency wins - usually) and that it doesn't break on a badly-written __eq__ (not an issue here and more of a paranoia anyway, but since there are no downsides, why not?).

answered Aug 30, 2011 at 19:30

user395760

2 Comments

user906780 Over a year ago

bad answer... I check my facts before posting, None without quotes doesn't work for me

user395760 Over a year ago

@Kingpin: That's a strong claim, you'll have to support it with code. None is a builtin and globally available, and group is documented to return None when the group didn't match.

Andrew Clark · Accepted Answer · 2011-08-30 20:01:10Z

1

I would suggest something like this:

>>> redfirst, othercolorfirst = zip(*(m.groups() for m in regex.finditer(string)))
>>> redfirst
(None, 'yellow', 'green')
>>> othercolorfirst
('blue', None, None)
>>> filter(None, redfirst)
('yellow', 'green')
>>> filter(None, othercolorfirst)
('blue',)
>>> print "\n".join(filter(None, redfirst))
yellow
green

answered Aug 30, 2011 at 20:01

Andrew Clark

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