Reduce resolution of array through summation

Question

If I have an array like this:

a = np.array([[ 1, 2, 3, 4],
              [ 5 ,6, 7, 8],
              [ 9,10,11,12],
              [13,14,15,16]])

I want to 'change the resolution', and end up with a smaller array, (say 2 rows by 2 cols, or 2 rows by 4 cols, etc.). I want this resolution change to happen through summation. I need this to work with large arrays, the number of rows, cols of the smaller array will always be a factor of the larger array.

Reducing the above array to a 2 by 2 array would result in (which is what I want):

[[ 14.  22.]
 [ 46.  54.]]

I have this function that does it fine:

import numpy as np

def shrink(data, rows, cols):
    shrunk = np.zeros((rows,cols))
    for i in xrange(0,rows):
        for j in xrange(0,cols):
            row_sp = data.shape[0]/rows
            col_sp = data.shape[1]/cols
            zz = data[i*row_sp : i*row_sp + row_sp, j*col_sp : j*col_sp + col_sp]
            shrunk[i,j] = np.sum(zz)
    return shrunk

print shrink(a,2,2)
print shrink(a,2,1)
#correct output:
[[ 14.  22.]
 [ 46.  54.]]
[[  36.]
 [ 100.]]

I've had a long look through the examples, but can't seem to find anything that helps.

Is there a faster way to do this, without needing the loops?

Answer 1

With your example:

a.reshape(2,2,2,2).sum(axis=1).sum(axis=2)

returns:

array([[14, 22],
       [46, 54]])

Now let's create a general function…

def shrink(data, rows, cols):
    return data.reshape(rows, data.shape[0]/rows, cols, data.shape[1]/cols).sum(axis=1).sum(axis=2)

works for your examples:

In [19]: shrink(a, 2,2)
Out[19]: 
array([[14, 22],
       [46, 54]])

In [20]: shrink(a, 2,1)
Out[20]: 
array([[ 36],
       [100]])