How to average list elements in Python

You can try this:

L=[[], [1.3342713788981633], [1.3342713788981633], [0.40896852187708455,1,3]]

L_avg = [ sum(l)/len(l) if len(l) > 0 else [] for l in L ]

print(L_avg)

Explanation: this iterates over all the sublists in L and calculates the average as the sum of the elements over the number of elements. The if-else is necessary to handle the case when the sublist is empty, which would give a division by 0.

You only be able to do this averaging if all your list members have values inside, in your case the first is empty, hence the error. It expected 2 parameters and got 0.

You can filter out the ones with len zero:

L_avg = [sum(item)/len(item) if len(item)>0 else []  for item in L] 

If you think about how to do this in a for loop it’ll make it easier to understand a comprehension.

for sl in L:
    print(sum(sl)/len(sl) if sl > 0 else [])

So the comprehension is:

[sum(sl)/len(sl) if sl > 0 else [] for sl in L]

How about

L_avg=[sum(sublist)/len(sublist) if sublist else [] for sublist in L]

Try this code:

L_avg = [sum(l)/len(l) if len(l)>0 else [] for l in L] 

it would probably make more sense to return None instead of [] when the list on computing the average is empty:

L_avg = [sum(l)/len(l) if len(l)>0 else None for l in L]