Difference between two time intervals in Python

Question

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. I'm using Python 2.6.5 on Windows XP.

Thanks.

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?

Answer 1

Yes, definitely datetime is what you need here. Specifically, the strptime function, which parses a string into a time object.

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

Answer 2

Structure that represent time difference in Python is called timedelta. If you have start_time and end_time which represents time or 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.

Answer 3

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).

Answer 4

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.

Answer 5

This site say to do:

import datetime as dt
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')
diff = (end_dt - start_dt) 
diff.seconds/60 

This forum uses time.mktime()