Adding Together Timedeltas In Python

So I have this list: [datetime.timedelta(0, 1800), datetime.timedelta(0, 1800), datetime.timedelta(0, 1800), datetime.timedelta(0, 1800)] Collectively that is 2:00 hours. I'm tryi

Solution 1:

The naive approach would be to take the seconds from each object and sum them

>>>a = [datetime.timedelta(0, 1800)] * 4>>>printsum([d.seconds for d in a])
7200
>>>printsum([d.seconds for d in a]) / 60.0 / 60.0
2.0

but this is not as robust as Haidro's solution:

importoperatorreduce(operator.add, a)

This results in a timedelta object with the correct delta that you can use however you want.

Solution 2:

The obvious way to sum anything number-like and addable in Python is with the sum function:

>>>dts = [datetime.timedelta(0, 1800), datetime.timedelta(0, 1800), datetime.timedelta(0, 1800), datetime.timedelta(0, 1800)]>>>sum(dts, start=datetime.timedelta(0))
datetime.timedelta(0, 9, 933279)

(For most number-like types, you don't even need the start value, because they know how to add themselves to 0. timedelta explicitly does not allow this, to avoid accidentally mixing dimensionless intervals—e.g., you don't want to add timeout_in_millis to a timedelta…)

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Whenever you're using reduce with operator.add for number-like values, you're probably doing it wrong.

If you don't provide an initial argument to reduce, it will do the wrong thing with an empty list, raising a TypeError instead of returning the 0 value. (The sum of no numbers is 0; the sum of no timedeltas is timedelta(0); etc.)

And if you do provide an initial argument, then it's just a more verbose, more complicated, and slower way to write sum. Compare:

functools.reduce(operator.add, a, foo(0))
sum(a, foo(0))