How to regroup or reconstruct connected elements in a binary image into squares

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Akakan-Abasi Okon 2021-8-29

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https://ww2.mathworks.cn/matlabcentral/answers/1442734-how-to-regroup-or-reconstruct-connected-elements-in-a-binary-image-into-squares

评论： Akakan-Abasi Okon 2021-9-1

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Hello, please I need help with this.

I have a binary image(50x50).

The matrix was randomly generated therefore making the image noisy. My aim is to regroup or reconstruct connected elements of same value together into different square shapes(not necessarily a perfect square), the edges of the square doesn't have to be smooth since the matrix is randomly generated. I did the following but I want to reconstruct all same elements into squares.

I = imread('Fig1.jpg');  
BW = imbinarize(I);  % Convert image to binary matrix.
L = bwlabeln(BW, 26) % Label connected elements.

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KALYAN ACHARJYA 2021-8-30

After seeing the image, I request you to rethink the question? What is your actual objective?

Akakan-Abasi Okon 2021-8-31

Hi @KALYAN ACHARJYA, Sorry I didn't respond on time. I was thinking of a better way to put the question. The objective here is to remove the scattered pixels making the image noisy without distorting so much of the main squares/rectangle hidden in the noisy image. I initially thought regrouping them would help but it will be fine to remove them rather.

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Answer 1

Image Analyst 2021-8-30

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https://ww2.mathworks.cn/matlabcentral/answers/1442734-how-to-regroup-or-reconstruct-connected-elements-in-a-binary-image-into-squares#answer_777124

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Why are you using bwlabeln()? For one thing, that's for 3-D images, not 2-D like you have. And what do you mean "of the same value"? Originally, you only have two values 0 and 1. Do you simply want a square that is all white with as many white dots there are in your original image?

numDots = nnz(BW);
rows = ceil(sqrt(numDots));
whiteSquare = ones(rows, rows);
numBlack = rows * rows - numDots;
if numBlack >= 1
    whiteSquare(1:numBlack) = 0;
end

Of course if you label them each connected component will have it's own ID number. Not sure if you want 4 connected or 8 connected but I believe if you use 26 connected on a 2-D image you'll get 8 connected (if it doesn't throw an error).

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Walter Roberson 2021-9-1

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h = randi([-1 1], 100, 80)
h = 100×80
   0    -1     0     0    -1     0    -1    -1     1     1    -1     0     1    -1    -1     1     1    -1    -1     1     1     1     0     0     1     0    -1     1     0
   1     0     0     0     1     1     0     1    -1    -1     1     1     0     0    -1    -1     0     1     0    -1     1     1     1     0     0     1     0    -1     1
   1    -1     0    -1     0     0     1    -1     1     0    -1    -1     0     0     1     0     1     1    -1    -1     0     0     0     0    -1     0    -1     0     1
    -1     0     1     0     0     1    -1    -1     1     1    -1    -1     0    -1     1     1    -1     1    -1     1     1     0     1    -1     0    -1     0     1     1    -1
  -1     1     1     1     0     0    -1     1    -1    -1    -1    -1    -1     0    -1    -1    -1    -1    -1     1     0     0     1     1    -1    -1     1     1     0
  -1     1     1    -1     0     0     0    -1    -1     0     0     0    -1    -1    -1     0    -1     0     0     0     1     1    -1    -1     0    -1     0    -1     0
  -1     1     1     1     1    -1     0    -1     1     0    -1     0     0    -1    -1    -1    -1    -1    -1     1    -1    -1     1     0     1    -1     1     0    -1
   0    -1    -1    -1     1    -1    -1     0     1     0     1     0     1     1    -1     0     1     1    -1    -1     0     1     1     1    -1    -1    -1    -1     0
    -1    -1     1     0    -1    -1     1     1     0    -1     1     1     0     1     0    -1    -1    -1    -1    -1    -1     1     1     0    -1     1     0     0    -1     0
    -1     1     0     0    -1     1     1     0     0     0     0    -1    -1     1    -1     1     0     1     0     0     0     1     0     0     1    -1     0     0    -1     0
h(h==-1) = 0
h = 100×80
   0     0     0     0     0     0     0     0     1     1     0     0     1     0     0     1     1     0     0     1     1     1     0     0     1     0     0     1     0
   1     0     0     0     1     1     0     1     0     0     1     1     0     0     0     0     0     1     0     0     1     1     1     0     0     1     0     0     1
   1     0     0     0     0     0     1     0     1     0     0     0     0     0     1     0     1     1     0     0     0     0     0     0     0     0     0     0     1
   0     1     0     0     1     0     0     1     1     0     0     0     0     1     1     0     1     0     1     1     0     1     0     0     0     0     1     1     0
   0     1     1     1     0     0     0     1     0     0     0     0     0     0     0     0     0     0     0     1     0     0     1     1     0     0     1     1     0
   0     1     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     1     1     0     0     0     0     0     0     0
   0     1     1     1     1     0     0     0     1     0     0     0     0     0     0     0     0     0     0     1     0     0     1     0     1     0     1     0     0
   0     0     0     0     1     0     0     0     1     0     1     0     1     1     0     0     1     1     0     0     0     1     1     1     0     0     0     0     0
   0     1     0     0     0     1     1     0     0     1     1     0     1     0     0     0     0     0     0     0     1     1     0     0     1     0     0     0     0
   1     0     0     0     1     1     0     0     0     0     0     0     1     0     1     0     1     0     0     0     1     0     0     1     0     0     0     0     0

Looks okay to me?

Image Analyst 2021-9-1

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Don't use the while. It's not needed and confusing since h<0 is a matrix -- a lot of values -- not a single value. What Walter gave you is sufficient and correct.

h(h == -1) = 0;

Again, delete the while and end lines.

Akakan-Abasi Okon 2021-9-1