Detecting coordinates of line edges

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Pankaj
Pankaj el 30 de Nov. de 2020
Comentada: Pankaj el 1 de Dic. de 2020
I have a matrix containing 1s and 0s. The elements of this matrix are pixels of an image, where 1&0 represents black and white. The 1s form non-intersecting line segmennts. I want to know pixel coordinates of line edges. Following is an example matrix-
binaryImage= [0 1 0 0;...
0 0 1 0; ...
0 0 0 1; ...
0 0 0 0]
The desired output is-
[1, 2], [3, 4]
Note that there could be multiple line segments. Is Hough transform useful here?
Thanks
--------------------
Update: I am also attaching another example matrix and corrosponding `imagesc` for better visualization of problem.
  3 comentarios
Pankaj
Pankaj el 30 de Nov. de 2020
No, just the ends.
sushanth govinahallisathyanarayana
sushanth govinahallisathyanarayana el 30 de Nov. de 2020
Yes, Hough might be useful, since you can describe a line with 2 points, however, you need an approximate idea of the scales at which you want to look, and how the lines may be oriented.

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Respuesta aceptada

Image Analyst
Image Analyst el 30 de Nov. de 2020
Use bwmorph() to get an image of just the endpoints, then use find() to get the rows and columns:
endPointsImage = bwmorph(binaryImage, 'endpoints');
[rows, columns] = find(endPointsImage);
rows and columns will be a list of the rows and columns of each endpoints. They are synced up so for index N, rows(N) has the row of the Nth endpoint and columns(N) has the column of the Nth endpoint.
  6 comentarios
Image Analyst
Image Analyst el 30 de Nov. de 2020
Yes. But you have to label the binary image so as to know which endpoint belongs to which blob. Keep in mind that is the blob is shaped like an X then it would have 4 endpoints, or 3 if it were shaped like a Y. So you could have different numbers of endpoints for each blob. So you'd do something like
[labeledImage, numRegions] = bwlabel(binaryImage);
endPointsImage = bwmorph(binaryImage, 'endpoints');
[rows, columns] = find(endPointsImage);
endpoints = cell(numRegions, 1; % At least 1 endpoint for each region
endpointsPerLabel = zeros(numRegions, 1);
for k = 1 : length(rows)
blobLabel = labeledImage(rows(k), columns(k));
endpointsPerLabel(blobLabel) = endpointsPerLabel(blobLabel) + 1;
x = columns(k);
y = rows(k);
endpoints{blobLabel, endpointsPerLabel(blobLabel)} = [x, y];
end
If this doesn't work, re-read my last comment, especially the last sentence.
Pankaj
Pankaj el 1 de Dic. de 2020
Thanks!! This is exactly what I wanted. I have also modified my question, slightly.

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Más respuestas (1)

KSSV
KSSV el 30 de Nov. de 2020
A= [0 1 0 0;...
0 0 1 0; ...
0 0 0 1; ...
0 0 0 0] ;
idx = find(A) ;
[i,j] = ind2sub(size(A),idx) ;
[i j]
  3 comentarios
KSSV
KSSV el 30 de Nov. de 2020
The above ives the locations of 1 in the matrix A.
Image Analyst
Image Analyst el 30 de Nov. de 2020
Pankaj, the code I gave you in my Answer gives the endpoints only. Did you overlook it?

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