Obtain part of a string using regular expressions in Python

Question

I am dealing with strings that need to be converted into dates in Python. In a normal situation, my string would have %d/%m/%Y %H:%M:%S. For instance: 18/02/2013 09:21:14

However in some occasion I could obtain something like %d/%m/%Y %H:%M:%S:%ms, such as:06/01/2014 09:52:14:78

I would like to get rid of that ms bit but I need to figure out how. I have been able to create a regular expression which can test if the date matches:

    mydate = re.compile("^((((31\/(0?[13578]|1[02]))|((29|30)\/(0?[1,3-9]|1[0-2])))\/(1[6-9]|[2-9]\d)?\d{2})|(29\/0?2\/(((1[6-9]|[2-9]\d)?(0[48]|[2468][048]|[13579][26])|((16|[2468][048]|[3579][26])00))))|(0?[1-9]|1\d|2[0-8])\/((0?[1-9])|(1[0-2]))\/((1[6-9]|[2-9]\d)?\d{2})) (20|21|22|23|[0-1]?\d):[0-5]?\d:[0-5]?\d$")
    s = "06/01/2014 09:52:14:78"
    bool(mydate.match(s))

>>> False

However I do not know how to obtain only the interesting part, i.e 06/01/2014 09:52:14 Any ideas?

Why not just try to strptime the string? If it fails, rsplit to remove the last part and try again. — jonrsharpe
– jonrsharpe, Commented Nov 28, 2014 at 14:37

Kasravnd · Accepted Answer · 2014-11-28 14:43:34Z

2

You can use a positive lookbehind and re.sub():

>>> re.sub(r'(?<=\d{2}:\d{2}:\d{2}).*','','06/01/2014 09:52:14:78')
'06/01/2014 09:52:14'

Regular expression visualization

Debuggex Demo

edited Nov 28, 2014 at 14:43

answered Nov 28, 2014 at 14:38

Kasravnd

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nu11p01n73R · Accepted Answer · 2014-11-28 14:46:06Z

1

How about the re.sub function

>>> re.sub(r'( \d{2}(:\d{2}){2}).*$',r'\1','06/01/2014 09:52:14:78')
'06/01/2014 09:52:14'
>>> re.sub(r'( \d{2}(:\d{2}){2}).*$,r'\1','8/02/2013 09:21:14')
'8/02/2013 09:21:14'

( \d{2}(:\d{2}){2}) matcheshours:min:sec` saved in capture group 1
.*$ matches the milliseconds
r'\1' replaced with contents of the first caputre group

edited Nov 28, 2014 at 14:46

answered Nov 28, 2014 at 14:38

nu11p01n73R

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