How to calculate the time interval between two time strings

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I have two times, a start and a stop time, in the format of 10:33:26 (HH:MM:SS). I need the difference between the two times. I’ve been looking through documentation for Python and searching online and I would imagine it would have something to do with the datetime and/or time modules. I can’t get it to work properly and keep finding only how to do this when a date is involved.

Ultimately, I need to calculate the averages of multiple time durations. I got the time differences to work and I’m storing them in a list. I now need to calculate the average. I’m using regular expressions to parse out the original times and then doing the differences.

For the averaging, should I convert to seconds and then average?

Yes, definitely datetime is what you need here. Specifically, the datetime.strptime() method, which parses a string into a datetime object.

from datetime import datetime
s1 = '10:33:26'
s2 = '11:15:49' # for example
FMT = '%H:%M:%S'
tdelta = datetime.strptime(s2, FMT) - datetime.strptime(s1, FMT)

That gets you a timedelta object that contains the difference between the two times. You can do whatever you want with that, e.g. converting it to seconds or adding it to another datetime.

This will return a negative result if the end time is earlier than the start time, for example s1 = 12:00:00 and s2 = 05:00:00. If you want the code to assume the interval crosses midnight in this case (i.e. it should assume the end time is never earlier than the start time), you can add the following lines to the above code:

if tdelta.days < 0:
    tdelta = timedelta(

(of course you need to include from datetime import timedelta somewhere). Thanks to J.F. Sebastian for pointing out this use case.

Try this — it’s efficient for timing short-term events. If something takes more than an hour, then the final display probably will want some friendly formatting.

import time
start = time.time()

time.sleep(10)  # or do something more productive

done = time.time()
elapsed = done - start

The time difference is returned as the number of elapsed seconds.

Here’s a solution that supports finding the difference even if the end time is less than the start time (over midnight interval) such as 23:55:00-00:25:00 (a half an hour duration):

#!/usr/bin/env python
from datetime import datetime, time as datetime_time, timedelta

def time_diff(start, end):
    if isinstance(start, datetime_time): # convert to datetime
        assert isinstance(end, datetime_time)
        start, end = [datetime.combine(datetime.min, t) for t in [start, end]]
    if start <= end: # e.g., 10:33:26-11:15:49
        return end - start
    else: # end < start e.g., 23:55:00-00:25:00
        end += timedelta(1) # +day
        assert end > start
        return end - start

for time_range in ['10:33:26-11:15:49', '23:55:00-00:25:00']:
    s, e = [datetime.strptime(t, '%H:%M:%S') for t in time_range.split('-')]
    print(time_diff(s, e))
    assert time_diff(s, e) == time_diff(s.time(), e.time())



time_diff() returns a timedelta object that you can pass (as a part of the sequence) to a mean() function directly e.g.:

#!/usr/bin/env python
from datetime import timedelta

def mean(data, start=timedelta(0)):
    """Find arithmetic average."""
    return sum(data, start) / len(data)

data = [timedelta(minutes=42, seconds=23), # 0:42:23
        timedelta(minutes=30)] # 0:30:00
# -> datetime.timedelta(0, 2171, 500000) # days, seconds, microseconds

The mean() result is also timedelta() object that you can convert to seconds (td.total_seconds() method (since Python 2.7)), hours (td / timedelta(hours=1) (Python 3)), etc.

This site says to try:

import datetime as dt
start_dt = dt.datetime.strptime(start, '%H:%M:%S')
end_dt = dt.datetime.strptime(end, '%H:%M:%S')
diff = (end_dt - start_dt) 

This forum uses time.mktime()

Structure that represent time difference in Python is called timedelta. If you have start_time and end_time as datetime types you can calculate the difference using - operator like:

diff = end_time - start_time

you should do this before converting to particualr string format (eg. before start_time.strftime(…)). In case you have already string representation you need to convert it back to time/datetime by using strptime method.

I like how this guy does it —
Not sure if it has some cons.

But looks neat for me 🙂

from datetime import datetime
from dateutil.relativedelta import relativedelta

t_a =
t_b =

def diff(t_a, t_b):
    t_diff = relativedelta(t_b, t_a)  # later/end time comes first!
    return '{h}h {m}m {s}s'.format(h=t_diff.hours, m=t_diff.minutes, s=t_diff.seconds)

Regarding to the question you still need to use datetime.strptime() as others said earlier.

Try this

import datetime
import time
start_time ='%H:%M:%S')
end_time ='%H:%M:%S')
total_time=(datetime.datetime.strptime(end_time,'%H:%M:%S') - datetime.datetime.strptime(start_time,'%H:%M:%S'))
print total_time



import datetime as dt
from dateutil.relativedelta import relativedelta

start = "09:35:23"
end = "10:23:00"
start_dt = dt.datetime.strptime(start, "%H:%M:%S")
end_dt = dt.datetime.strptime(end, "%H:%M:%S")
timedelta_obj = relativedelta(start_dt, end_dt)

0 0 0 0 -47 -37

Both time and datetime have a date component.

Normally if you are just dealing with the time part you’d supply a default date. If you are just interested in the difference and know that both times are on the same day then construct a datetime for each with the day set to today and subtract the start from the stop time to get the interval (timedelta).

Take a look at the datetime module and the timedelta objects. You should end up constructing a datetime object for the start and stop times, and when you subtract them, you get a timedelta.

    import datetime
    day = int(input("day[1,2,3,..31]: "))
    month = int(input("Month[1,2,3,...12]: "))
    year = int(input("year[0~2020]: "))
    start_date =, month, day)
    day = int(input("day[1,2,3,..31]: "))
    month = int(input("Month[1,2,3,...12]: "))
    year = int(input("year[0~2020]: "))
    end_date =, month, day)
    time_difference = end_date - start_date
    age = time_difference.days
    print("Total days: " + str(age))

Concise if you are just interested in the time elapsed that is under 24 hours. You can format the output as needed in the return statement :

import datetime
def elapsed_interval(start,end):
    elapsed = end - start
    min,secs=divmod(elapsed.days * 86400 + elapsed.seconds, 60)
    hour, minutes = divmod(min, 60)
    return '%.2d:%.2d:%.2d' % (hour,minutes,secs)

if __name__ == '__main__':
    """ do your process """

you can use pendulum:

import pendulum

t1 = pendulum.parse("10:33:26")
t2 = pendulum.parse("10:43:36")

period = t2 - t1


would output:


Usually, you have more than one case to deal with and perhaps have it in a pd.DataFrame(data) format. Then:

import pandas as pd

df['duration'] = pd.to_datetime(df['stop time']) - pd.to_datetime(df['start time'])

gives you the time difference without any manual conversion.

Taken from Convert DataFrame column type from string to datetime.

If you are lazy and do not mind the overhead of pandas, then you could do this even for just one entry.

Here is the code if the string contains days also [-1 day 32:43:02]:

    (int(time.replace('-', '').split(' ')[0]) * 24) * 60 
    + (int(time.split(' ')[-1].split(':')[0]) * 60)
    + int(time.split(' ')[-1].split(':')[1])

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