Parsing your strings using strptime()
:
a = time.strptime('2013-10-05T01:21:07Z', '%Y-%m-%dT%H:%M:%SZ')
b = time.strptime('2013-10-05T01:21:16Z', '%Y-%m-%dT%H:%M:%SZ')
This will parse the given time strings as local times (setting daylight savings (DST) to automatic), and the results are time structs. These still reflect whether DST was explicitly off (0), on (1), or implicitly automatic (-1). Convert these to a float (seconds since 1970-01-01):
a = time.mktime(a)
b = time.mktime(b)
Then compute the difference (in seconds):
d = b - a
And convert them to days/hours/minutes/seconds:
days = int(d) / 86400
hours = int(d) / 3600 % 24
minutes = int(d) / 60 % 60
seconds = int(d) % 60
The last block only works properly for positive differences, so be careful not to swap the a
and b
;-)
But @J.F.Sebastian correctly pointed out that this might not be what you intended. It seems from the notation that your strings describe a UTC time, not a local time. For mere time differences this is relevant in case your time spans over a DST switch. In this case it would of course result in a time difference one hour too great or one hour too small (because UTC is always without DST).
To avoid this, you can set the DST flag from automatic (-1) to a fixed value (e. g. 0 for off) and use these values:
a = time.mktime(a[:-1] + (0,)) # switch DST to off
b = time.mktime(b[:-1] + (0,))
Or, also as @J.F.Sebastian pointed out, you could forget about the time
module and instead use datetime.datetime
which is unaware of the DST aspect:
a = datetime.datetime.strptime('2013-10-05T01:21:07Z', '%Y-%m-%dT%H:%M:%SZ')
b = datetime.datetime.strptime('2013-10-05T01:21:16Z', '%Y-%m-%dT%H:%M:%SZ')
Then the results are datetime
objects which can be subtracted directly to get a timedelta
object which represents such a time difference as you want it. Printing it will result in sth like 0:00:05
which might well be exactly what you are looking for.