Python – Regex to Split 1st Colon

pythonregex

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…

Best Answer

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

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

Related Solutions

Regex – How to validate phone numbers using regex

Better option... just strip all non-digit characters on input (except 'x' and leading '+' signs), taking care because of the British tendency to write numbers in the non-standard form +44 (0) ... when asked to use the international prefix (in that specific case, you should discard the (0) entirely).

Then, you end up with values like:

 12345678901
 12345678901x1234
 345678901x1234
 12344678901
 12345678901
 12345678901
 12345678901
 +4112345678
 +441234567890

Then when you display, reformat to your hearts content. e.g.

  1 (234) 567-8901
  1 (234) 567-8901 x1234

Python – How to split a list into evenly sized chunks

Here's a generator that yields the chunks you want:

def chunks(lst, n):
    """Yield successive n-sized chunks from lst."""
    for i in range(0, len(lst), n):
        yield lst[i:i + n]

import pprint
pprint.pprint(list(chunks(range(10, 75), 10)))
[[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
 [20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
 [30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
 [40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
 [50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
 [60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
 [70, 71, 72, 73, 74]]

If you're using Python 2, you should use xrange() instead of range():

def chunks(lst, n):
    """Yield successive n-sized chunks from lst."""
    for i in xrange(0, len(lst), n):
        yield lst[i:i + n]

Also you can simply use list comprehension instead of writing a function, though it's a good idea to encapsulate operations like this in named functions so that your code is easier to understand. Python 3:

[lst[i:i + n] for i in range(0, len(lst), n)]

Python 2 version:

[lst[i:i + n] for i in xrange(0, len(lst), n)]

Best Answer

Related Solutions

Regex – How to validate phone numbers using regex

Python – How to split a list into evenly sized chunks

Related Topic