Regex to Split 1st Colon

Question

I have a time in ISO 8601 ( 2009-11-19T19:55:00 ) which is also paired with a name commence. I'm trying to parse this into two. I'm currently up to here:

import re
sColon = re.compile('[:]')

aString = sColon.split("commence:2009-11-19T19:55:00")

Obviously this returns:

>>> aString
['commence','2009-11-19T19','55','00']

What I'd like it to return is this:

>>>aString
['commence','2009-11-19T19:55:00']

How would I go about do this in the original creation of sColon? Also, do you recommend any Regular Expression links or books that you have found useful, as I can see myself needing it in the future!

EDIT:

To clarify... I'd need a regular expression that would just parse at the very first instance of :, is this possible? The text ( commence ) before the colon can chance, yes...

Answer 1

Looks like you need .IndexOf(":"), then .Substring()?

Answer 2

>>> first, colon, rest = "commence:2009-11-19T19:55:00".partition(':')

>>> print (first, colon, rest)
('commence', ':', '2009-11-19T19:55:00')

Answer 3

You could put maximum split parameter in split function

>>> "commence:2009-11-19T19:55:00".split(":",1)
['commence', '2009-11-19T19:55:00']

Official Docs

S.split([sep [,maxsplit]]) -> list of strings

Return a list of the words in the string S, using sep as the
delimiter string.  If maxsplit is given, at most maxsplit
splits are done. If sep is not specified or is None, any
whitespace string is a separator and empty strings are removed
from the result.

Answer 4

@OP, don't do the unnecessary. Regex is not needed with what you are doing. Python has very good string manipulation methods that you can use. All you need is split(), and slicing. Those are the very basics of Python.

>>> "commence:2009-11-19T19:55:00".split(":",1)
['commence', '2009-11-19T19:55:00']
>>>