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.

Generation of the submatrices is too awkward that is difficult to read and can only be used when the filter is 3*3
The size of the varient submatrices seems to be too big, since it is approximately 9 folds bigger than the original matrix.
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

Asked By: Allosteric

||

Source

Answer 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]]

Answered By: Crispin

Answer 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)

Answered By: Daniel F

Answer 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

Answered By: Dan Erez

Answer 4

https://laurentperrinet.github.io/sciblog/posts/2017-09-20-the-fastest-2d-convolution-in-the-world.html

Check out all the convolution methods and their respective performances here.
Also, I found the below code snippet to be simpler.

from numpy.fft  import fft2, ifft2
def np_fftconvolve(A, B):
    return np.real(ifft2(fft2(A)*fft2(B, s=A.shape)))

Answered By: Naman Bansal

Convolve2d just by using Numpy

Question:

What I have done

Answers: