I want to find end to end distance of this skeleton image
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filename = 'https://www.mathworks.com/matlabcentral/answers/uploaded_files/1045155/insect.jpg';
k = imread(filename);
k = rgb2gray(k);
%imshow(insect);
%applyig threshold to convert the image into binary so that backgroung can
%be separated
BI = k < 160;
%imshow(BI)
BI = bwareaopen(BI,40);
%imshow(BI)
%finding the skeleton of image
k = bwskel(BI);
%imshow(k);
[a b]=size(k);
output=zeros(a,b);
for i=2:a-1
for j=2:b-1
ws=[k(i-1,j-1),k(i-1,j),k(i-1,j+1),k(i,j-1),k(i,j)...
,k(i,j+1),k(i+1,j-1),k(i+1,j),k(i+1,j+1)];
kg=sum(ws);
if(k(i,j)==1 && kg==2)
output(i,j)=1;
end
end
end
figure;
output=imdilate(output,strel('diamond',5));
imshow(output+k);
i have this code with which i highlighted the end points but dont know how to go further
2 comentarios
Walter Roberson
el 25 de Jun. de 2022
It is not obvious to me which are the endpoints, unless you mean the endpoints of every one of the (sometimes very short) line segments ?
Respuestas (1)
Image Analyst
el 25 de Jun. de 2022
Do you mean the sum of all white pixels to the neighbor pixels? Have you tried pdist2, and extract all pairs just 1 or sqrt(2) pixels apart, sum them, and divide by 2? Something like (untested):
[y, x] = find(k); % Find coordinates.
xy = [x(:), y(:)]; % Combine into matrix.
distances = pdist2(xy, xy); % Find distance of every point to every other point.
mask = distances < mean([sqrt(2), sqrt(3)]); % Find pixels that are adjacent or on nearest diagonal ONLY.
sumOfDistances = sum(distances(mask)) / 2 % Divide by 2 since we get the sum for each endpoint so it's counted twice.
9 comentarios
Walter Roberson
el 26 de Jun. de 2022
I wonder if you could describe why that distance is significant? I would have expected it to be more significant to trace along the curve, to measure the length of the creature, instead of calculating something about how much it is curled up at the moment?
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