parse date/time

[bb8]

i need to parse date/time stamps of this format:

Oct 4 13:27:52 2003

i also need to handle leap years.

what can i use for this (besides regexp which seems complicated to write)?

[Larz60+]

import time
import datetime

dt = 'Oct 4 13:27:52 2003'
dtobj = datetime.datetime.strptime(dt, '%b %d %H:%M:%S %Y')
print(dtobj)
timestamp = int(time.mktime(dtobj.timetuple()))
print(timestamp)

results:

Output:
datetime.datetime(2003, 10, 4, 13, 27, 52)
1065288472

[bb8]

what if i have something else too on the line (like, 4 5 6 <date>)?

[Larz60+]

you will have to figure out where the date starts and parse from there.

if you knew that the dates all begin with a valid month
This is a hack, but works:

>>> import calendar
>>> import time
>>> import datetime
>>> monlst = []
>>> for mon in range(1, 13):
...     monlst.append(calendar.month_abbr[mon])
...
>>> dt = '4 5 6 Oct 4 13:27:52 2003'
>>> sdt = dt.strip().split()
>>> for word in sdt:
...     if word in monlst:
...         idx = dt.index(word)
...         dt = dt[idx:]
...
>>> dtobj = datetime.datetime.strptime(dt, '%b %d %H:%M:%S %Y')
>>> print(dtobj)
2003-10-04 13:27:52
>>> timestamp = int(time.mktime(dtobj.timetuple()))
>>> print(timestamp)
1065288472
>>>

[bb8]

i also need the first 3 numbers

[Larz60+]

Put in a bit of effort.
How do you think this can be done?

[bb8]

i was thinking of first getting the numbers, then doing the date/time

[Larz60+]

In the code that I showed you,
the input is split into a list.
guess what the first three cells of the list contain.

[bb8]

ignore this, it's okay

when i have something like this: "4 5 6 Oct 4 13:27:00 2003", notice the second is 0, strptime() produces (2003, 10, 4, 13, 27) and not (2003, 10, 4, 13, 27, 0)