Get average of times

Question

Given the following list:

times = [datetime.timedelta(0, 1, 256273), datetime.timedelta(0, 0, 910417), datetime.timedelta(0, 0, 388175)]

How would I get the average of times? Doing the following gives me an error:

avg = (float(sum(times)) / len(times))
TypeError: unsupported operand type(s) for +: 'int' and 'datetime.timedelta'

Answer 1

There are two problems here:

• sum() starts out the summing with an integer 0. You cannot add a timedelta() object to an integer, so you get an error:

  >>> import datetime
  >>> times = [datetime.timedelta(0, 1, 256273), datetime.timedelta(0, 0, 910417), datetime.timedelta(0, 0, 388175)]
  >>> sum(times)
  Traceback (most recent call last):
    File "<stdin>", line 1, in <module>
  TypeError: unsupported operand type(s) for +: 'int' and 'datetime.timedelta'
  

• Even if you could just use sum() outright, you cannot convert a timedelta() object to a float() by passing it to the float() function:

  >>> float(datetime.timedelta())
  Traceback (most recent call last):
    File "<stdin>", line 1, in <module>
  TypeError: float() argument must be a string or a number
  

You need to tell sum() to start with an empty timedelta (start at timedelta(0)) and use the timedelta.total_seconds() function to get a floating point number:

>>> sum(times, datetime.timedelta())
datetime.timedelta(0, 2, 554865)
>>> sum(times, datetime.timedelta()).total_seconds()
2.554865
>>> sum(times, datetime.timedelta()).total_seconds() / len(times)
0.8516216666666666

You can omit the .total_seconds() part and get another timedelta() object:

>>> sum(times, datetime.timedelta()) / len(times)
datetime.timedelta(0, 0, 851621)

which is going to be more accurate. You can always call .total_seconds() after the division:

>>> (sum(times, datetime.timedelta()) / len(times)).total_seconds()
0.851621

Answer 2

Another option is to use list comprehension to get a list of floats, which can be summed:

avg = sum([t.total_seconds() for t in times]) / len(times)

# avg == 0.851621666667

But as Martijn said, it is more accurate to sum timedelta objects together, starting from an empty one.