Want to find a way of doing an average of multiple lists

Question

Say we create a list like so in python:

[[1, 2, 3], [1, 3, 4], [2, 4, 5]]

And then I want to take 1+1+2 and divide by 3, giving me the average for that element and store in a new list. I want to do that again for the second elements and lastly for the third. How would one do it succinctly? (I cannot think of a way other than multiple loops.)

The output should be a new list [(1+1+2), (2+3+4), (3+4+5)]

Thanks so much!

Solution 1

Averages:

>>> data = [[1, 2, 3], [1, 3, 4], [2, 4, 5]]
>>> from __future__ import division
>>> [sum(e)/len(e) for e in zip(*data)]
 [1.3333333333333333, 3.0, 4.0]

Sums:

>>> data = [[1, 2, 3], [1, 3, 4], [2, 4, 5]]
>>> [sum(e) for e in zip(*data)]
 [4, 9, 12]

---

• zip

returns a list of tuples, where the i-th tuple contains the i-th element from each of the argument sequences or iterables.

• Unpacking argument lists

when the arguments are already in a list or tuple but need to be unpacked for a function call requiring separate positional arguments ... write the function call with the *-operator to unpack the arguments out of a list or tuple.

>>> data
 [[1, 2, 3], [1, 3, 4], [2, 4, 5]]

>>> zip(*data)
 [(1, 1, 2), (2, 3, 4), (3, 4, 5)]

Solution 2

>>> l =  [[1, 2, 3], [1, 3, 4], [2, 4, 5]]
>>> zip(*l)
[(1, 1, 2), (2, 3, 4), (3, 4, 5)]
>>> def average(nums, default=float('nan')):
...   return sum(nums) / float(len(nums)) if nums else default
... 
>>> [average(n) for n in zip(*l)]
[2.0, 2.6666666666666665, 3.6666666666666665]

Comments

[abarnert]

I think you want zip(*l) in your last line. Otherwise, you're just averaging each sublist (so you get 2, 2-2/3, and 3-2/3 instead of 1-1/3, 3, and 4).