不同工作日的两个时间戳之间的区别？

Question

I've the following dates:

Tue 19:00-20:00
Wed 04:25-15:14

How can i get the difference between Tue 20:00 and Wed 04:25 ?

I've tried to transform it in datetime objects but since i don't have the day of the month or a year both objects will get the same date:

d1 = datetime.strptime("Tue 20:00", "%a %H:%M")
d2 = datetime.strptime("Wed 04:25", "%a %H:%M")

• d1 is: datetime.datetime(1900, 1, 1, 20, 0)
• d2 is: datetime.datetime(1900, 1, 1, 4, 25)

The difference in minutes should be 505 but in this case will be -935.0 because they have the same date, and so they differ by almost 16 hours and not about 8 and a half:

diff = (d2 - d1).total_seconds() / 60  # -935.0

Answer 1

datetime.strptime() can't do anything else, because the day of the week is not enough to form a date. You'd need a week number and a year for a week-day to make any sense. If you need to get the relative difference between two dates that are assumed to be in the same week , then any week number and year would do, eg:

arbitrary_iso_week = "2020-25 "

d1 = datetime.strptime(arbitrary_iso_week + "Tue 20:00", "%G-%V %a %H:%M")
d2 = datetime.strptime(arbitrary_iso_week + "Wed 04:25", "%G-%V %a %H:%M")

This makes use of the %G and %V formatters for ISO 8601 week numbers .

This produces datetime objects with a more meaningful date releationship:

>>> arbitrary_iso_week = "2020-25 "
>>> d1 = datetime.strptime(arbitrary_iso_week + "Tue 20:00", "%G-%V %a %H:%M")
>>> d2 = datetime.strptime(arbitrary_iso_week + "Wed 04:25", "%G-%V %a %H:%M")
>>> d1
datetime.datetime(2020, 6, 16, 20, 0)
>>> d2
datetime.datetime(2020, 6, 17, 4, 25)
>>> (d2 - d1).total_seconds() / 60
505.0

The actual dates don't matter here, just their difference in days. d1 is now earlier, not later, than d2 .

You may have to take into account that week days are circular, in that you could have an earlier date on, say, Friday, and a later date on Monday. If you can assume that the first value must be earlier than the second, then the solution is simple. Just subtract 7 days from d1 if it is later than d2 :

# if the first date must come before the second, then it must be one week earlier.
if d1 > d2:
    d1 -= timedelta(days=7)