python regex optional non capturing space

Question

I want to match:

!argument1 [argument2], where argument2 is optional. I have it figured out using a second optional group, but I can't get it to ignore the space between the arguments.

regex="!([a-z]) ?\W(.)"

Matches: "!test case"

"!test "

Not matching: "!test"

I tried to throw in an '?(: )' in the regex to make a non capturing optional group containing the space but to no avail, resulting in the following: regex="!([a-z])?(: )?\W(.)"

The result is 'raise error, v # invalid expression'

some tries usin the python shell:

>>>  regex="!([a-z]*) ?\W(.*)"
>>> data="!hello dear"
>>> re.search(regex, data).groups()
('hello', 'dear')
>>> data="!hello "
>>> re.search(regex, data).groups()
('hello', '')
>>> data="!hello"
>>> re.search(regex, data).groups()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: 'NoneType' object has no attribute 'groups'

You've only made the space optional. The second group is mandatory. — user2357112
– user2357112, Commented Jul 28, 2014 at 10:59
user3218114: my wildcard is missing due to formatting. user2357112: Well yes, but I dont mind if its empty. Is it better if I add the (optional) whitespace to the secondary group and when I want to make use of that data I simply strip the whitespace? — user135361
– user135361, Commented Jul 28, 2014 at 11:16
Why u need a space before \W \W (means non word char) so you can simply use !([a-z]+)\W?(.*) optional non word char followed by second group which is captured then — gaw
– gaw, Commented Jul 28, 2014 at 11:32

gaw · Accepted Answer · 2014-07-28 11:38:22Z

1

The regex in the comments should fullfill your requirements !([a-z]+) ?(.*) You require a ! followed by a sequence of characters (your first argument) then there might be a space and another capturing group which could be empty(so missing) or contain any characters after the space

answered Jul 28, 2014 at 11:38

gaw

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Comments

Mardoxx · Accepted Answer · 2014-07-28 12:03:04Z

0

Overcomplicated.

import re
e = re.compile("(!?\w+)")
s = "!test ing code"
args = e.findall(s)[0][0]
print args

['!test', 'ing', 'code']

then just check if args[0][0] == "!"

I assume this is for a bot of some sort, probably an IRC one, could look at phenny.

Even easier would be to see if your commands starts with ! and strip it before branching to the regex parser. Then you can just capture \w+

answered Jul 28, 2014 at 12:03

Mardoxx

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