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.
datetime(1,1,1,0,0,0) instead of date.today().
datetime exit, since exit is a built-in function.
combine method in my datetime module: AttributeError: 'module' object has no attribute 'combine'
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.
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.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)
timedelta conceptually represents a span of time, not a time of day.
timedelta conceptually represents a span of time and not a point in time, which makes it the correct thing to use here, as we are implicitly talking about the time that has passed (a delta of time) since the start of the day (a point in time).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())
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.
datetime.min + deltadatetime.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
timedelta accepts negative(-) time values. so it could be simple as below.
Answer (single line, without date combine)
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)
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.
datetime.time objects a and b are from the same day, and that b > a, then the operation b - a has perfect meaning.
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?
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())
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)
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.
I think the reason for Time objects being defined this way is that the result of timeA - timeB depends on the date, as we could hit a daylight saving boundary.
