Calculating Mean of arrays with different lengths

Question

Is it possible to calculate the mean of multiple arrays, when they may have different lengths? I am using numpy. So let's say I have:

numpy.array([[1, 2, 3, 4, 8],    [3, 4, 5, 6, 0]])
numpy.array([[5, 6, 7, 8, 7, 8], [7, 8, 9, 10, 11, 12]])
numpy.array([[1, 2, 3, 4],       [5, 6, 7, 8]])

Now I want to calculate the mean, but ignoring elements that are 'missing' (Naturally, I can not just append zeros as this would mess up the mean)

Is there a way to do this without iterating through the arrays?

PS. These arrays are all 2-D, but will always have the same amount of coordinates for that array. I.e. the 1st array is 5 and 5, 2nd is 6 and 6, 3rd is 4 and 4.

An example:

np.array([[1, 2],    [3, 4]])
np.array([[1, 2, 3], [3, 4, 5]])
np.array([[7],       [8]])

This must give

(1+1+7)/3  (2+2)/2   3/1
(3+3+8)/3  (4+4)/2   5/1

And graphically:

[1, 2]    [1, 2, 3]    [7]
[3, 4]    [3, 4, 5]    [8]

Now imagine that these 2-D arrays are placed on top of each other with coordinates overlapping contributing to that coordinate's mean.

Answer 1

numpy.ma.mean allows you to compute the mean of non-masked array elements. However, to use numpy.ma.mean, you have to first combine your three numpy arrays into one masked array:

import numpy as np
x = np.array([[1, 2], [3, 4]])
y = np.array([[1, 2, 3], [3, 4, 5]])
z = np.array([[7], [8]])

arr = np.ma.empty((2,3,3))
arr.mask = True
arr[:x.shape[0],:x.shape[1],0] = x
arr[:y.shape[0],:y.shape[1],1] = y
arr[:z.shape[0],:z.shape[1],2] = z
print(arr.mean(axis = 2))

yields

[[3.0 2.0 3.0]
 [4.66666666667 4.0 5.0]]