将嵌套列表值的平均值放入新列表中

Question

I have the following list:

x = [(27.3703703703704, 2.5679012345679, 5.67901234567901,
      6.97530864197531, 1.90123456790123, 0.740740740740741,
      0.440136054421769, 0.867718446601942),
     (25.2608695652174, 1.73913043478261, 6.07246376811594,
      7.3768115942029, 1.57971014492754, 0.710144927536232,
      0.4875, 0.710227272727273)]

I'm looking for a way to get the average of each of the lists nested within the main list, and create a new list of the averages. So in the case of the above list, the output would be something like:

[[26.315],[2.145],[5.87],etc...]

I would like to apply this formula regardless of the amount of lists nested within the main list.

Answer 1

I assume your list of tuples of one-element lists is looking for the sum of each unpacked element inside the tuple, and a list of those options. If that's not what you're looking for, this won't work.

result = [sum([sublst[0] for sublst in tup])/len(tup) for tup in x]

EDIT to match changed question

result = [sum(tup)/len(tup) for tup in x]

EDIT to match your even-further changed question

result = [[sum(tup)/len(tup)] for tup in x]

Answer 2

An easy way to acheive this is:

means = []     # Defines a new empty list
for sublist in x:     # iterates over the tuples in your list
    means.append([sum(sublist)/len(sublist)])    # Put the mean of the sublist in the means list

This will work no matter how many sublists are in your list. I would advise you read a bit on list comprehensions: https://docs.python.org/2/tutorial/datastructures.html

Answer 3

It looks like you're looking for the zip function:

[sum(l)/len(l) for l in zip(*x)]

zip combines a collection of tuples or lists pairwise, which looks like what you want for your averages. then you just use sum()/len() to compute the average of each pair.

*x notation means pass the list as though it were individual arguments, ie as if you called: zip(x[0], x[1], ..., x[len(x)-1])

Answer 4

r = [[sum(i)/len(i)] for i in x]