Convolve2d just by using Numpy

Question

I am studying image-processing using Numpy and facing a problem with filtering with convolution.

I would like to convolve a gray-scale image. (convolve a 2d Array with a smaller 2d Array)

Does anyone have an idea to refine my method ?

I know that scipy supports convolve2d but I want to make a convolve2d only by using Numpy.

What I have done

First, I made a 2d array the submatrices.

a = np.arange(25).reshape(5,5) # original matrix

submatrices = np.array([
     [a[:-2,:-2], a[:-2,1:-1], a[:-2,2:]],
     [a[1:-1,:-2], a[1:-1,1:-1], a[1:-1,2:]],
     [a[2:,:-2], a[2:,1:-1], a[2:,2:]]])

the submatrices seems complicated but what I am doing is shown in the following drawing.

Next, I multiplied each submatrices with a filter.

conv_filter = np.array([[0,-1,0],[-1,4,-1],[0,-1,0]])
multiplied_subs = np.einsum('ij,ijkl->ijkl',conv_filter,submatrices)

and summed them.

np.sum(np.sum(multiplied_subs, axis = -3), axis = -3)
#array([[ 6,  7,  8],
#       [11, 12, 13],
#       [16, 17, 18]])

Thus this procudure can be called my convolve2d.

def my_convolve2d(a, conv_filter):
    submatrices = np.array([
         [a[:-2,:-2], a[:-2,1:-1], a[:-2,2:]],
         [a[1:-1,:-2], a[1:-1,1:-1], a[1:-1,2:]],
         [a[2:,:-2], a[2:,1:-1], a[2:,2:]]])
    multiplied_subs = np.einsum('ij,ijkl->ijkl',conv_filter,submatrices)
    return np.sum(np.sum(multiplied_subs, axis = -3), axis = -3)

However, I find this my_convolve2d troublesome for 3 reasons.

1. Generation of the submatrices is too awkward that is difficult to read and can only be used when the filter is 3*3
2. The size of the varient submatrices seems to be too big, since it is approximately 9 folds bigger than the original matrix.
3. The summing seems a little non intuitive. Simply said, ugly.

Thank you for reading this far.

Kind of update. I wrote a conv3d for myself. I will leave this as a public domain.

def convolve3d(img, kernel):
    # calc the size of the array of submatracies
    sub_shape = tuple(np.subtract(img.shape, kernel.shape) + 1)

    # alias for the function
    strd = np.lib.stride_tricks.as_strided

    # make an array of submatracies
    submatrices = strd(img,kernel.shape + sub_shape,img.strides * 2)

    # sum the submatraces and kernel
    convolved_matrix = np.einsum('hij,hijklm->klm', kernel, submatrices)

    return convolved_matrix

Solution 1

You could generate the subarrays using as_strided:

import numpy as np

a = np.array([[ 0,  1,  2,  3,  4],
       [ 5,  6,  7,  8,  9],
       [10, 11, 12, 13, 14],
       [15, 16, 17, 18, 19],
       [20, 21, 22, 23, 24]])

sub_shape = (3,3)
view_shape = tuple(np.subtract(a.shape, sub_shape) + 1) + sub_shape
strides = a.strides + a.strides

sub_matrices = np.lib.stride_tricks.as_strided(a,view_shape,strides)

To get rid of your second "ugly" sum, alter your einsum so that the output array only has j and k. This implies your second summation.

conv_filter = np.array([[0,-1,0],[-1,5,-1],[0,-1,0]])
m = np.einsum('ij,ijkl->kl',conv_filter,sub_matrices)

# [[ 6  7  8]
#  [11 12 13]
#  [16 17 18]]

Solution 2

Cleaned up using as_strided and @Crispin 's einsum trick from above. Enforces the filter size into the expanded shape. Should even allow non-square inputs if the indices are compatible.

def conv2d(a, f):
    s = f.shape + tuple(np.subtract(a.shape, f.shape) + 1)
    strd = numpy.lib.stride_tricks.as_strided
    subM = strd(a, shape = s, strides = a.strides * 2)
    return np.einsum('ij,ijkl->kl', f, subM)

Solution 3

You can also use fft (one of the faster methods to perform convolutions)

from numpy.fft import fft2, ifft2
import numpy as np

def fft_convolve2d(x,y):
    """ 2D convolution, using FFT"""
    fr = fft2(x)
    fr2 = fft2(np.flipud(np.fliplr(y)))
    m,n = fr.shape
    cc = np.real(ifft2(fr*fr2))
    cc = np.roll(cc, -m/2+1,axis=0)
    cc = np.roll(cc, -n/2+1,axis=1)
    return cc

• https://gist.github.com/thearn/5424195
• you must pad the filter to be the same size as image ( place it in the middle of a zeros_like mat.)

cheers, Dan

Comments

[Marijn van Vliet]

thank you for providing drawings of the matrices :) If I understand correctly, you want tips on how to make your solution more elegant?

[Allosteric]

Glad it helps! Yes. I would be grateful if you can provide me tips to overcome the 3 problems written in the very last lines.

[Allosteric]

I should add that the 3 points are rather arranged in a priority order. The first one is quite important for me and the last one seems kinda trivial. I will also be glad if there are other problems and refinements about it.

[Andreas K.]

Isn't the second drawing (after the equality sign) wrong? Shouldn't each submatrix be multiplied (element-wise) with the filter, and then the elements of each of the resulting submatrices summed?

[Allosteric]

@AndyK They will produce the same result.

[NaN]

if a_s is the strided array and filter is your laplacian like filter, then try... np.sum(a_s*filter, axis=(2,3)) if indeed your answer is array([[ 6, 7, 8], [11, 12, 13], [16, 17, 18]])

[NaN]

simplify further... see my comment below... np.sum(a_s * filter, axis=(2,3)) if indeed your answer is array([[ 6, 7, 8], [11, 12, 13], [16, 17, 18]]) ... where a_s is strided array and filter is the 3x3 filter

[Allosteric]

Thank you for the tip. I am trying this myself right now. Maybe trivial but I believe that the name filter is not appropriate because it is a built-in function of python.

[Daniel F]

Not sure why that works @NaN since it is certainly not doing what the problem asked - but it does, even for arbitrary a matrices

[Crispin]

You can do the summation directly in the Einstein summation. See answer

[Allosteric]

In this question, isn't it better to write, sub_shape = conv_filter.shape ?

[Daniel F]

It doesn't seem that it would necessarily be the case. conv_filter.shape should be a.shape - sub_shape + 1. It just so happens in your example a.shape = [5, 5] and sub_shape = (3, 3) so conv_filter.shape == sub_shape is true but not strictly enforced. If you want it to be enforced, then you should set sub_shape = a.shape - conv_filter.shape + 1

[Allosteric]

Got it! So view_shape = conv_filter.shape?

[Daniel F]

well view_shape should be a 4D tuple, so view.shape = tuple(conv_filter.shape) + tuple(sub_shape), so you still need to find sub_shape

[Allosteric]

at least at numpy v12, a.shape and f.shape is a tuple so s should be tuple(np.subtract(a.shape, f.shape)+1), I think.

[Daniel F]

True! Still used to shape giving an ndarray. Fixed that.

[michaelsnowden]

Follow-up question here: stackoverflow.com/questions/45580013/…

[sirgogo]

Hi @DanielF, is there a generalization of this that works for RGB? Not super familiar with einsum notation, so an idea how to generalize would be awesome.

[Daniel F]

@sirgogo, I suggest asking a seperate question linking this answer. There are people watching numpy way better at multidimensional convlcutions than I am.

[sirgogo]

@DanielF, thanks, here is the more general version of what I'm trying to do: [stackoverflow.com/questions/50239641/…

[Foad S. Farimani]

I have a similar issue for higher dimensions. I was wondering if you could be so kind to take a look at my question here

[Cris Luengo]

It is certainly not efficient to use the FFT for convolution if the kernel has only 9 elements. This is only useful for large kernels.

[Dan Erez]

True. It depends on the kernel and image size but the threshold for fft to outperform is very low. It also depends on your computer architecture but for most generic use-cases fft is the go to in high performance libraries.

[Naman Bansal]

The following article compares the performance of various image convolution methods, and thus, we can conclude that the FFT method of Numpy is the fastest method to perform convolution. laurentperrinet.github.io/sciblog/posts/…

[Cris Luengo]

This code is not correct. It will shift the output image. The larger the kernel, the larger the shift.

[Naman Bansal]

Can you elaborate on how will it shift the output image?

[Cris Luengo]

stackoverflow.com/questions/54877892/…

[Michael Kopp]

Is using einsum("ij,klij", ...) sufficient/better? Using the code from this answer with input of different shape, I got operands could not be broadcast together with remapped shapes, using "ij,klij" seems to work without issues.

[OuttaSpaceTime]

is the kernel flipped here or is it without flipping?