need the average from a list of timedelta objects

Question

I have created a list of timedelta objects and i need to get the average of this list. when i try to do

return (sum(delta_list)/(len(delta_list)-1))

i get

TypeError: unsupported operand type(s) for +: 'int' and 'datetime.timedelta'

I am new at working with python's datetime classes. I also would like to know how to get this average into a format like days: hours: mins: i don't need anything smaller than mins.

`Thanks in advanced.`. Don't thank in advance and forget to thank later. Learn to accept your answers. — user225312, Feb 15 '11 at 16:19
What's wrong with the code you've posted? Does it get an error? Please show the error you get? — S.Lott, Feb 15 '11 at 16:22
I forgot to paste it, I edited the question to include the error. — Cory, Feb 15 '11 at 16:26

Jochen Ritzel · Accepted Answer · 2011-02-15T16:39:54.030

31

sum wants a starting value, which is 0 by default, but 0 can't be added to a timedelta so you get the error.

You just have to give sum a timedelta() as the start value:

# this is the average
return sum(delta_list, timedelta()) / len(delta_list)

To print it out you can do this:

print str(some_delta)

If you want something customized you can get some_delta.days and some_delta.seconds but you have to calculate everything between.

edited Feb 15 '11 at 16:39

answered Feb 15 '11 at 16:27

Jochen Ritzel

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@cferbs: Yes it's correct, to get the average you do `sum( .. ) / len( .. )` – Jochen Ritzel Feb 15 '11 at 16:35
In Python 3.3 make sure you use the '//' operator: – Boden Garman Dec 14 '12 at 05:00

score 4 · Answer 2 · answered Feb 15 '11 at 16:31

First of all, sum adds all of the elements from the list to an initial value, which is by default 0 (an integer with value of zero). So to be able to use sum on a list of timedelta's, you need to pass a timedelta(0) argument - sum(delta_list, timedelta(0)).

Secondly, why do you want to subtract one from the length of the list? If a list is of length 1 you'll get a ZeroDivisionError.

Having fixed the above you'll end up with a timedelta object being an average. How to convert that to a human-readable format I'm leaving for you as an exercise. The datetime module documentation should answer all of the possible questions.

I realized it was wrong. I got confused from other sample code from http://stackoverflow.com/questions/179716/average-difference-between-dates-in-python — Cory, Feb 15 '11 at 16:38

David Webb · Answer 3 · 2011-02-15T16:36:32.690

The sum() function needs a start value to add all the items in the iterable to. It defaults to 0 which is why you're getting the error about adding a timedelta to an int.

To fix this, just pass a timedelta object as the second argument to sum:

(Creating a timedelta with no arguments creates one corresponding to a zero time difference.)

Also, the mean of a set of items is usually the sum of items divided by the number of items so you don't need to subtract 1 from len(delta_list).

These changes mean we can remove some of the brackets from your statement.

So this gives you:

return sum(delta_list,timedelta()) / len(delta_list)

score 1 · Answer 4 · answered Feb 15 '11 at 16:30

A datetime.timedelta object has attributes to access the days, microseconds, and seconds. You can reference these easily, for example:

>>> import datetime
>>> daybefore = datetime.timedelta(days=1)
>>> dayminus2 = datetime.timedelta(days=2, minutes=60)
>>> daybefore.days
1
>>> dayminus2.days, dayminus2.seconds
(2, 3600)

To get minutes you're going to have to divide seconds by 60.

need the average from a list of timedelta objects

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