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subtract two times in python

Posted by: admin November 29, 2017 Leave a comment


I have two datetime.time values, exit and enter and I want to do something like:

duration = exit - enter

However, I get this error:

TypeError: unsupported operand type(s) for -: 'datetime.time' and 'datetime.time

How do I do this correctly? One possible solution is converting the time variables to datetime variables and then subtruct, but I’m sure you guys must have a better and cleaner way.


Try this:

from datetime import datetime, date

datetime.combine(date.today(), exit) - datetime.combine(date.today(), enter)

combine builds a datetime, that can be subtracted.



from datetime import datetime, date

duration = datetime.combine(date.min, end) - datetime.combine(date.min, beginning)

Using date.min is a bit more concise and works even at midnight.

This might not be the case with date.today() that might return unexpected results if the first call happens at 23:59:59 and the next one at 00:00:00.

import datetime

def diff_times_in_seconds(t1, t2):
    # caveat emptor - assumes t1 & t2 are python times, on the same day and t2 is after t1
    h1, m1, s1 = t1.hour, t1.minute, t1.second
    h2, m2, s2 = t2.hour, t2.minute, t2.second
    t1_secs = s1 + 60 * (m1 + 60*h1)
    t2_secs = s2 + 60 * (m2 + 60*h2)
    return( t2_secs - t1_secs)

# using it
diff_times_in_seconds( datetime.datetime.strptime( "13:23:34", '%H:%M:%S').time(),datetime.datetime.strptime( "14:02:39", '%H:%M:%S').time())


datetime.time does not support this, because it’s nigh meaningless to subtract times in this manner. Use a full datetime.datetime if you want to do this.


instead of using time try timedelta:

from datetime import timedelta

t1 = timedelta(hours=7, minutes=36)
t2 = timedelta(hours=11, minutes=32)
t3 = timedelta(hours=13, minutes=7)
t4 = timedelta(hours=21, minutes=0)

arrival = t2 - t1
lunch = (t3 - t2 - timedelta(hours=1))
departure = t4 - t3

print(arrival, lunch, departure)


I had similar situation as you and I ended up with using external library called arrow.

Here is what it looks like:

>>> import arrow
>>> enter = arrow.get('12:30:45', 'HH:mm:ss')
>>> exit = arrow.now()
>>> duration = exit - enter
>>> duration
datetime.timedelta(736225, 14377, 757451)


The python timedelta library should do what you need. A timedelta is returned when you subtract two datetime instances.

import datetime
dt_started = datetime.datetime.utcnow()

# do some stuff

dt_ended = datetime.datetime.utcnow()
print((dt_ended - dt_started).total_seconds())