Average values in two Numpy arrays

>>> import numpy as np
>>> old_set = [[0, 1], [4, 5]]
>>> new_set = [[2, 7], [0, 1]]
>>> (np.array(old_set) + np.array(new_set)) / 2.0
array([[1., 4.],
[2., 3.]])

You can create a 3D array containing your 2D arrays to be averaged, then average along axis=0 using np.mean or np.average (the latter allows for weighted averages):

np.mean( np.array([ old_set, new_set ]), axis=0 )

This averaging scheme can be applied to any (n)-dimensional array, because the created (n+1)-dimensional array will always contain the original arrays to be averaged along its axis=0.

Using numpy.average

Also numpy.average can be used with the same syntax:

import numpy as np
a = np.array([np.arange(0,9).reshape(3,3),np.arange(9,18).reshape(3,3)])
averaged_array = np.average(a,axis=0)

The advantage of numpy.average compared to numpy.mean is the possibility to use also the weights parameter as an array of the same shape:

weighta = np.empty((3,3))
weightb = np.empty((3,3))
weights = np.array([weighta.fill(0.5),weightb.fill(0.8) ])
np.average(a,axis=0,weights=weights)

If you use masked arrays consider also using numpy.ma.average because numpy.average don't deal with them.

As previously said, your solution does not work because of the nested lists (2D matrix). Staying away from numpy methods, and if you want to use nested for-loops, you can try something like:

old_set = [[0, 1], [4, 5]]
new_set = [[2, 7], [0, 1]]


ave_set = []
for i in range(len(old_set)):
row = []
for j in range(len(old_set[0])):
row.append( ( old_set[i][j] + new_set[i][j] ) / 2 )
ave_set.append(row)
print(ave_set) # returns [[1, 4], [2, 3]]