Convert datetime.time into datetime.timedelta in Python 3.4

我们两清 提交于 2019-12-17 16:58:51

问题


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.


回答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.




回答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.



来源:https://stackoverflow.com/questions/35241643/convert-datetime-time-into-datetime-timedelta-in-python-3-4

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!