subtract two times in python

Question

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.

Solution 1

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.

Solution 2

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.

Solution 3

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)

Solution 4

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())

Solution 5

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.

View more solutions

Comments

[swalog]

If you know from your domain that two datetime.time objects a and b are from the same day, and that b > a, then the operation b - a has perfect meaning.

[Akavall]

To @IgnacioVazquez-Abrams point, one could use, say,datetime(1,1,1,0,0,0) instead of date.today().

[naught101]

Even if they aren't the same day, it still makes fine sense. At least as much sense as arctan(0) = (0, pi, 2pi, ...), but we just don't care about any of those values after the first. So, 4:00 - 20:00 is 8:00 - it's also (32:00, 56:00, ... ), but who cares?

[Ignacio Vazquez-Abrams]

Unless it's one of those cases that doesn't.

[orokusaki]

Don't name a datetime exit, since exit is a built-in function.

[mtoloo]

using python 2.7, I don't have combine method in my datetime module: AttributeError: 'module' object has no attribute 'combine'

[gruszczy]

mtoloo: There is datetime module and it has a datetime object inside. The object inside has combine method. If you are simply importing datetime (like this: import datetime), then what you need to do later is this datetime.datetime.combine.

[Sean]

Furthermore, Python doesn't even stay consistent. If a - b is meaningless for two time objects, then how it that a > b or a < b behaves as it does? The comparison is already assuming same day, otherwise it is completely broken. Since the assumption is made for comparison, it is entirely consistent to make the same assumption for difference operations.

[Sean]

or date.min -> datetime.date(1, 1, 1)

[ramwin]

There is a very rare case when you call this function at some time like 23:59:59:9999. In that case, the date.today() before and the date.today() later will return a different value. It would be better to give the value date.today() to a variable.

[ramwin]

Agree with you.

[aj.toulan]

Saying it's meaningless is subjective, but I see where you're coming from. Adding two times together seems to be a meaningless representation to me. I think subtracting two time values should return a timedelta. If that timedelta happens to be negative, so be it. What I want to know is why python can't add or subtract a timedelta from a time to return a new time value.

[Pushpak Dagade]

Thanks. This is very clean and better than all the answers here - stackoverflow.com/questions/3096953/…

[dejanualex]

It is possible to convert the timedelta object back to datetime ? I mean we have difference_delta .seconds(), is there any way to get back a datetime or time object ? Thx

[sourcedelica]

Yes. datetime.min + delta

[kleptog]

Here you are just demonstrating subtracting complete datetime objects, not times-of-day as in the original question