在Python 3.4中将datetime.time转换为datetime.timedelta

Question

I am trying to convert two "durations", however I am currently receiving a TypeError due to one being a datetime.timedelta and one being a datetime.time :

TypeError: unorderable types: datetime.time() <= datetime.timedelta()

What is an efficient way to convert a datetime.time to a datetime.timedelta ?

I have checked the docs and there is no built-in method for conversion between these two types.

Answer 1

datetime.time() is not a duration , it is a point in a day. If you want to interpret it as a duration, then convert it to a duration since midnight:

datetime.combine(date.min, timeobj) - datetime.min

Demo:

>>> from datetime import datetime, date, time
>>> timeobj = time(12, 45)
>>> datetime.combine(date.min, timeobj) - datetime.min
datetime.timedelta(0, 45900)

You may need to examine how you get the datetime.time() object in the first place though, perhaps there is a shorter path to a timedelta() from the input data you have? Don't use datetime.time.strptime() for durations, for example.

Answer 2

Here's one solution I've found, though it's not necessarily efficient:

import datetime
x = datetime.timedelta(hours=x.hour, minutes=x.minute, seconds=x.second, microseconds=x.microsecond)

Where x is a datetime.time object.