在Python中减去两次

I have two datetime.time values, exit and enter and I want to do something like:

duration = exit - enter

However, I get this error:

TypeError: unsupported operand type(s) for -: ‘datetime.time’ and ‘datetime.time

How do I do this correctly? One possible solution is converting the time variables to datetime variables and then subtruct, but I’m sure you guys must have a better and cleaner way.

Try this:

from datetime import datetime, date

datetime.combine(date.today(), exit) - datetime.combine(date.today(), enter)

combine builds a datetime, that can be subtracted.

Use:

from datetime import datetime, date

duration = datetime.combine(date.min, end) - datetime.combine(date.min, beginning)

Using date.min is a bit more concise and works even at midnight.

This might not be the case with date.today() that might return unexpected results if the first call happens at 23:59:59 and the next one at 00:00:00.

instead of using time try timedelta:

from datetime import timedelta

t1 = timedelta(hours=7, minutes=36)
t2 = timedelta(hours=11, minutes=32)
t3 = timedelta(hours=13, minutes=7)
t4 = timedelta(hours=21, minutes=0)

arrival = t2 - t1
lunch = (t3 - t2 - timedelta(hours=1))
departure = t4 - t3

print(arrival, lunch, departure)

datetime.time does not support this, because it’s nigh meaningless to subtract times in this manner. Use a full datetime.datetime if you want to do this.

You have two datetime.time objects so for that you just create two timedelta using datetime.timedetla and then substract as you do right now using “-” operand. Following is the example way to substract two times without using datetime.

enter = datetime.time(hour=1)  # Example enter time
exit = datetime.time(hour=2)  # Example start time
enter_delta = datetime.timedelta(hours=enter.hour, minutes=enter.minute, seconds=enter.second)
exit_delta = datetime.timedelta(hours=exit.hour, minutes=exit.minute, seconds=exit.second)
difference_delta = exit_delta - enter_delta

difference_delta is your difference which you can use for your reasons.

The python timedelta library should do what you need. A timedelta is returned when you subtract two datetime instances.

import datetime
dt_started = datetime.datetime.utcnow()

# do some stuff

dt_ended = datetime.datetime.utcnow()
print((dt_ended - dt_started).total_seconds())

import datetime

def diff_times_in_seconds(t1, t2):
    # caveat emptor - assumes t1 & t2 are python times, on the same day and t2 is after t1
    h1, m1, s1 = t1.hour, t1.minute, t1.second
    h2, m2, s2 = t2.hour, t2.minute, t2.second
    t1_secs = s1 + 60 * (m1 + 60*h1)
    t2_secs = s2 + 60 * (m2 + 60*h2)
    return( t2_secs - t1_secs)

# using it
diff_times_in_seconds( datetime.datetime.strptime( "13:23:34", '%H:%M:%S').time(),datetime.datetime.strptime( "14:02:39", '%H:%M:%S').time())

datetime.time can not do it – But you could use datetime.datetime.now()

start = datetime.datetime.now()
sleep(10)
end = datetime.datetime.now()
duration = end - start

I had similar situation as you and I ended up with using external library called arrow.

Here is what it looks like:

>>> import arrow
>>> enter = arrow.get('12:30:45', 'HH:mm:ss')
>>> exit = arrow.now()
>>> duration = exit - enter
>>> duration
datetime.timedelta(736225, 14377, 757451)