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
댓글 수: 2
Walter Roberson
2022년 6월 25일
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 ?
답변 (1개)
Image Analyst
2022년 6월 25일
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.
댓글 수: 9
Walter Roberson
2022년 6월 26일
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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