How to parse a string like '2016-08-01 13:39:00+05:30' into datetime object of python

Question

I am getting below error while trying to do this

from datetime import datetime 

time1 = '2016-08-01 13:39:00+05:30'
x = datetime.strptime(time1, '%Y-%m-%d %H:%M:%S %z')
print(x)

Error is ...

ValueError: time data '2016-08-01 13:39:00+05:30' does not match format '%Y-%m-%d %H:%M:%S %z'

didn't worked ...ValueError: time data '2016-08-01 13:39:00+05:30' does not match format '%Y-%m-%d %H:%M:%S%z' — Sudheer1990, Aug 01 '16 at 08:48
`%z` does not even work on my computer.. What version of Python are you using? Anyway, [it's not recommended](http://stackoverflow.com/a/12282040/5018771) to use timezones in strptime — AdrienW, Aug 01 '16 at 08:53
read more here: http://stackoverflow.com/questions/20194496/iso-to-datetime-object-z-is-a-bad-directive — Juraj Bezručka, Aug 01 '16 at 08:57

score 1 · Answer 1 · edited Sep 27 '17 at 04:54

If you are using Python 2 or early versions of Python 3 (3.0 and 3.1), you can use the dateutil library for converting a string to a timezone aware object.

The code to do this is simple:

>>> import dateutil.parser
>>> dt = dateutil.parser.parse('2016-08-01 13:39:00+05:30')
>>> dt
datetime.datetime(2016, 8, 1, 13, 39, tzinfo=tzoffset(None, 19800))

If you are using Python 3.2 or later, the %z option has been added as a formatting option when parsing a date. You can accomplish this task without using dateutil in these versions by doing this:

>>> import datetime
>>> dt = datetime.datetime.strptime('2016-08-01 13:39:00+0530', "%Y-%m-%d %H:%M:%S%z")
>>> dt
datetime.datetime(2016, 8, 1, 13, 39, tzinfo=datetime.timezone(datetime.timedelta(0, 19800)))

Unfortunately, you do have to strip the colon (:) from the offset for this to work as expected.

score 0 · Answer 2 · answered Aug 01 '16 at 08:56

0

this works in python 3.4, conforms to the datetime documentation

from datetime import datetime

time1 = '2016-08-01 13:39:00+0530'
x = datetime.strptime(time1, '%Y-%m-%d %H:%M:%S%z')
print(x)

gives

2016-08-01 13:39:00+05:30

answered Aug 01 '16 at 08:56

Jean-François Fabre

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@DombiSzabolcs `%z` uses the form +HHMM or -HHMM – Natecat Aug 01 '16 at 09:04
to make it work you should remove the last ':' `time2 = ':'.join(time1.split(':')[:-1]) + time1.split(':')[-1]` – Juraj Bezručka Aug 01 '16 at 09:05
Yes, I altered input data without thinking, Isn't it strange that the printed value cannot be parsed back by `datetime`? isn't there a bug or improvement for python? – Jean-François Fabre Aug 01 '16 at 09:12

score 0 · Answer 3 · answered Sep 28 '16 at 21:06

Consider using dateparser:

>>> dateparser.parse('2016-08-01 13:39:00+05:30')
datetime.datetime(2016, 8, 1, 13, 39)
>>> dateparser.parse('2016-08-01 13:39:00+05:30', settings={'TO_TIMEZONE': 'UTC'})
datetime.datetime(2016, 8, 1, 8, 9)

How to parse a string like '2016-08-01 13:39:00+05:30' into datetime object of python

3 Answers3