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.

Answer 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.

Answer 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.

Answer 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)

Answer 4

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.

Answer 5

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())

Answer 6

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

Answer 7

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.

Answer 8

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())

Answer 9

timedelta accepts negative(-) time values. so it could be simple as below.

Answer (single line)

datetime.timedelta(hours=exit.hour-enter.hour, minutes=exit.minute-enter.minute)

---

Run test

import datetime

enter = datetime.time(hour=1, minute=30)
exit = datetime.time(hour=2, minute=0)
duration = datetime.timedelta(hours=exit.hour-enter.hour, minutes=exit.minute-enter.minute)

>>> duration
datetime.timedelta(seconds=1800)

Answer 10

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)

Answer 11

import time
from datetime import datetime

def calcTime(enter,exit):
    format="%H:%M:%S"
    #Parsing the time to str and taking only the hour,minute,second 
    #(without miliseconds)
    enterStr = str(enter).split(".")[0]
    exitStr = str(exit).split(".")[0]
    #Creating enter and exit time objects from str in the format (H:M:S)
    enterTime = datetime.strptime(enterStr, format)
    exitTime = datetime.strptime(exitStr, format)
    return exitTime - enterTime

enter = datetime.today().time()
#Sleeping for 5 seconds before initializing the exit variable
time.sleep(5)
exit = datetime.today().time()
duration = calcTime(enter,exit)
print(f"Duration is {duration} (Hours:Minutes:Seconds)")
#Output: Duration is 0:00:05 (Hours:Minutes:Seconds)

If it is helpful, you can use the calcTime function as shown.