Question

Extract an 8 × 8 patch from the image. To access to the (i, j)th pixel of the image, you can type Y(i,j). To access the an 8×8 patch at pixel (i, j), you can do Y(i:i+7, j:j+7) for some index i and j. Extract all the available 8 × 8 patches from the image, and store them as a 64 × K matrix where K is the number of patches. The following code will be useful.

for i=1:M-8

for j=1:N-8

z = Y(i+[0:7], j+[0:7]);

... % other steps; your job.

end

end

Here, M and N are the number of rows and columns of the image, respectively. No need to worry about the boundary pixels; just drop them.

Hint: Use the command reshape in MATLAB (and Python) to turn an 8 × 8 patch into a 64 × 1 column vector. Submit the first 2 columns and the last 2 columns of your data matrix.

Answer 1

import cv2
import numpy as np
import matplotlib.pyplot as plt
file_path='pic\logo.jpg'        #location of file from current directory
img = cv2.imread(file_path)     
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)                #since the image is an 
                                                #3d(r,g,b) array but for this question
#cv2.imshow('image',gray)               ## covert into a gray image(2d) by taking only
                                         ## first channel.
m,n=gray.shape                          ##shape of your pic
size=(m-7)*(n-7)                        ##total number of distinct matrix can be formed
arr=np.empty((64,size))   
k=0

for i in range(m-7):
    for j in range(n-7):
        patch=gray[i:i+8,j:j+8]
        patch=patch.reshape(64,1)
        arr[:,k]=patch[:,0]    
        k=k+1
print(arr.shape)  
cv2.waitKey(0)      ###pause screen until key press
cv2.destroyAllWindows()     

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