CODEWITHSUNDEEP
pythonnumpyimage-processingconv-neural-networkconvolution
Suhail Gupta
Suhail Gupta

Reputation: 23256

Correctly using the numpy's convolve with an image

I was watching Andrew Ng's videos on CNN and wanted to to convolve a 6 x 6 image with a 3 x 3 filter. The way I approached this with numpy is as follows:

image = np.ones((6,6))
filter = np.ones((3,3))

convolved = np.convolve(image, filter)

Running this gives an error saying:

ValueError: object too deep for desired array

I could comprehend from the numpy documentation of convolve on how to correctly use the convolve method.

Also, is there a way I could do a strided convolutions with numpy?

Upvotes: 3

Views: 6625

Answers (1)

gmds
gmds

Reputation: 19885

np.convolve function, unfortunately, only works for 1-D convolution. That's why you get an error; you need a function that allows you to perform 2-D convolution.

However, even if it did work, you actually have the wrong operation. What is called convolution in machine learning is more properly termed cross-correlation in mathematics. They're actually almost the same; convolution involves flipping the filter matrix followed by performing cross-correlation.

To solve your problem, you can look at scipy.signal.correlate (also, don't use filter as a name, as you'll shadow the inbuilt function):

from scipy.signal import correlate

image = np.ones((6, 6))
f = np.ones((3, 3))

correlate(image, f)

Output:

array([[1., 2., 3., 3., 3., 3., 2., 1.],
       [2., 4., 6., 6., 6., 6., 4., 2.],
       [3., 6., 9., 9., 9., 9., 6., 3.],
       [3., 6., 9., 9., 9., 9., 6., 3.],
       [3., 6., 9., 9., 9., 9., 6., 3.],
       [3., 6., 9., 9., 9., 9., 6., 3.],
       [2., 4., 6., 6., 6., 6., 4., 2.],
       [1., 2., 3., 3., 3., 3., 2., 1.]])

This is the standard setting of full cross-correlation. If you want to remove elements which would rely on the zero-padding, pass mode='valid':

from scipy.signal import correlate

image = np.ones((6, 6))
f = np.ones((3, 3))

correlate(image, f, mode='valid')

Output:

array([[9., 9., 9., 9.],
       [9., 9., 9., 9.],
       [9., 9., 9., 9.],
       [9., 9., 9., 9.]])

Upvotes: 5

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