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I was told that applying average pooling to a matrix M is equivalent to dropping the high frequencies components of the Fourier representation of M. With average pooling I mean 2 by 2 average pooling as visualized in this image:

enter image description here

I wanted to verify this and see how this worked using numpy. So I wrote a naive implementation of average pooling and copied a function to neatly display matrices from here:

def prettyPrintMatrix(m):
    s = [['{:.3f}'.format(e) for e in row] for row in m]
    lens = [max(map(len, col)) for col in zip(*s)]
    fmt = '\t'.join('{{:{}}}'.format(x) for x in lens)
    table = [fmt.format(*row) for row in s]
    print '\n'.join(table)

def averagePool(im):
    imNew = np.empty((im.shape[0] /2, im.shape[1]/2))
    for i in range(imNew.shape[0]):
        for j in range(imNew.shape[1]):
            imNew[i,j] = np.average(im[(2*i):(2*i+2), (2*j):(2*j+2)])
    return imNew

Now to test what happens to the Fourier coefficients I ran the following code:

M = np.random.random((8,8))
Mpooled = averagePool(M)

# print the original M
print('original M:')
prettyPrintMatrix(M)

# print Fourier coefficients of regular matrix
print('Fourier of M')
prettyPrintMatrix(np.fft.fft2(M))

# print Fourier coefficients of pooled matrix
print('Fourier of the pooled M')
prettyPrintMatrix(np.fft.fft2(Mpooled))

An example output for this would be:

original M:
0.849   0.454   0.231   0.605   0.375   0.842   0.533   0.954
0.489   0.097   0.990   0.199   0.572   0.262   0.299   0.634
0.477   0.052   0.429   0.670   0.323   0.458   0.459   0.954
0.984   0.884   0.620   0.657   0.352   0.765   0.897   0.642
0.179   0.894   0.835   0.710   0.916   0.544   0.968   0.557
0.253   0.197   0.813   0.450   0.936   0.165   0.169   0.712
0.677   0.544   0.507   0.107   0.733   0.334   0.056   0.171
0.356   0.639   0.580   0.517   0.763   0.401   0.771   0.219

Fourier of M
34.680+0.000j   -0.059-0.188j   0.076+1.227j    -1.356+1.515j   2.101+0.000j    -1.356-1.515j   0.076-1.227j    -0.059+0.188j
-1.968-1.684j   2.125-0.223j    2.277+1.442j    1.629-0.795j    -0.141+1.460j   0.694-2.363j    -0.627+0.971j   -0.847-2.094j
3.496+2.808j    -1.099+1.260j   0.921-0.814j    2.499+0.283j    -1.048-1.206j   -3.228+2.435j   -2.934+0.030j   0.386-0.015j 
0.451-0.301j    0.791-0.143j    -0.463-0.031j   1.841+0.032j    -1.979-1.066j   1.344-1.229j    3.487-1.297j    2.105-2.455j 
0.111+0.000j    0.166+1.317j    0.946-0.016j    0.587-0.443j    -2.710+0.000j   0.587+0.443j    0.946+0.016j    0.166-1.317j 
0.451+0.301j    2.105+2.455j    3.487+1.297j    1.344+1.229j    -1.979+1.066j   1.841-0.032j    -0.463+0.031j   0.791+0.143j 
3.496-2.808j    0.386+0.015j    -2.934-0.030j   -3.228-2.435j   -1.048+1.206j   2.499-0.283j    0.921+0.814j    -1.099-1.260j
-1.968+1.684j   -0.847+2.094j   -0.627-0.971j   0.694+2.363j    -0.141-1.460j   1.629+0.795j    2.277-1.442j    2.125+0.223j 

Fourier of the pooled M
8.670+0.000j    -0.180+0.019j   -0.288+0.000j   -0.180-0.019j
-0.228-0.562j   0.487+0.071j    0.156+0.638j    -0.049-0.328j
0.172+0.000j    -0.421-0.022j   -0.530+0.000j   -0.421+0.022j
-0.228+0.562j   -0.049+0.328j   0.156-0.638j    0.487-0.071j

Now I would expect Fourier coefficients of the pooled matrix to somehow be related to the low frequency Fourier coefficients of the original matrix. However the only relation I see is the very lowest frequency (top left), which is exactly 4 times smaller after pooling. My question is now: is there a relation between the Fourier coefficients before and after pooling and if so, what is it?

dimpol
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    An averaging filter like this is a very poor low pass filter, so you can't expect to see an ideal frequency response. You will see a general reduction in the higher frequency components, but it won't be uniform. – Paul R May 31 '17 at 15:12
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    Ok, the problem lies in the theory, not with the implementation. Nobody posted an answer. So if you post this as an answer, I will accept it. – dimpol Jun 02 '17 at 14:35
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    No problem - comment converted to answer below. – Paul R Jun 02 '17 at 15:16

1 Answers1

2

An averaging filter is a very crude low pass filter, so you can't expect to see an ideal frequency response. You will see a general reduction in the higher frequency components, but it won't be uniform. Averaging is typically used when efficiency is a consideration (since the only coefficients are implicit 1 and 0 values, so only addition is required), or when accuracy is not important. Otherwise a proper low-pass filter should be used.

Paul R
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