Mean of list of unaqual lists

Question

I try to get a list that include the mean of all of (i)th element in all lists of a list.To explain more, if i have the next list:

A=[[1,2],  [2,3], [1,2,3], [3,4,5], [2,2,1]].

I want to have [a,b,c] with:

a = (1+2+1+3+2)/5
b = (2+3+2+4+2)/5
c = (3+5+1)/3

Any help please!

Answer 1

use itertools.zip_longest to zip values together. The extra difficulty is that the length of the data varies. zip_longest inserts None as a fill value. We have to create a filtered list, and count only those items:

import itertools

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

tuples = itertools.zip_longest(*A)
for t in tuples:
    s = [x for x in t if x is not None]
    print(sum(s)/len(s))

result:

1.8
2.6
3.0

EDIT: one-liner (maybe a bit more complex, using lambda to compute mean):

result = [(lambda x : sum(x)/len(x))([x for x in t if x is not None]) for t in itertools.zip_longest(*A)]

result: [1.8, 2.6, 3.0]

(Python 2 users have to replace zip_longest by izip_longest )

Answer 2

You can try with itertools.izip_longest() like so if you are on Python2:

from itertools import izip_longest
from __future__ import division

A=[[1,2],  [2,3], [1,2,3], [3,4,5], [2,2,1]]
a,b,c = [sum(filter(None, c))/len(filter(None,c)) for c in izip_longest(*A)]

print a
print b
print c

Output:

1.8
2.6
3.0

Answer 3

Use itertools.zip_longest to generate the sublists, and statistics.mean to calculate the average for each group, disregarding the None s:

from itertools import zip_longest
from statistics import mean

averages = list(map(lambda l: mean(i for i in l if i != None), zip_longest(*A)))

With your example:

>>> A = [[1, 2], [2, 3], [1, 2, 3], [3, 4, 5], [2, 2, 1]]
>>> from itertools import zip_longest
>>> from statistics import mean
>>> list(map(lambda l: mean(i for i in l if i != None), zip_longest(*A)))
[1.8, 2.6, 3]