I have an object of type datetime.time. How do I convert this to an int representing its duration in seconds? Or to a string, which I can then convert to a second representation by splitting?
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1Duration assumes you have two datetime.time objects and a difference of seconds between them. Are you trying to calculate the difference in seconds between two datetime.time objects? – BoboDarph Jun 29 '17 at 10:55
4 Answers
You can calculate it by yourself:
from datetime import datetime
t = datetime.now().time()
seconds = (t.hour * 60 + t.minute) * 60 + t.second
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You need to convert your datetime.time object into a datetime.timedelta to be able to use total_seconds() function.
It will return a float rather than an int as asked in the question but you can easily cast it.
>>> from datetime import datetime, date, time, timedelta
>>> timeobj = time(12, 45)
>>> t = datetime.combine(date.min, timeobj) - datetime.min
>>> isinstance(t, timedelta)
# True
>>> t.total_seconds()
45900.0
Links I've be inspired by:
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If your object is supposed to represent a duration, you should use a datetime.timedelta instead of a datetime.time.
datetime.time objects are meant to represent a time of the day.
datetime.timedelta objects are meant to represent a duration, and have a total_seconds() method that does exactly what you want.
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As @Owl Max proposed the easiest way to have an integer representation of a time is to use a timedelta. Although, I would like to share a different way to construct the timedelta.
A useful one-liner I like to use is:
import datetime
t = datetime.time(10, 0, 5)
int(datetime.timedelta(hours=t.hour, minutes=t.minute, seconds=t.second).total_seconds())
(ps. Normally this would be a comment)
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