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Any suggestions to calculate the threshold of pixel intensity (the dip represents low intensity)

조회 수: 1 (최근 30일)
imshow(matrix(:,:,1))
[x ,y]= ginput(2);
I = matrix(:,:,1);
for i=1:size(matrix,3)
I=matrix(:,:,i);
test = improfile(I(:,:,1), [x(1) x(2)] , [y(1) y(2)]);
end
plot(test);
x1= xline(20, 'color', 'r');
x2=xline(29, 'color', 'b');
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Image Analyst
Image Analyst 2022년 8월 1일
What exactly are you looking for? The index where the intensity profile crosses 40? The halfway point from the max intensity to the min intensity?

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Image Analyst
Image Analyst 2022년 8월 1일
Try this:
[cx, xy, profile] = improfile(matrix(:,:,1), [x(1), x(2)] , [y(1), y(2)]);
threshold = (max(profile) + min(profile)) / 2; % Half way point in intensity.
indexLeft = find(profile >= threshold, 1, 'first')
xLeft = cx(indexLeft);
yLeft = cy(indexLeft);
indexRight = find(profile >= threshold, 1, 'last')
xRight = cx(indexRight);
yRight = cy(indexRight);
If you have any more questions, then attach your data and code to read it in with the paperclip icon after you read this:
  댓글 수: 3
Image Analyst
Image Analyst 2022년 8월 1일
Give the complete code, not a snippet. And tell me where you clicked on the image.
Yes, I renamed your badly-named "test" variable to a more descriptively -named "profile". But that is not a problem. It will still work.
I don't know what "calculate the 'aperture'" means. I took the profile and found both the x and y location for the threshold at both the left side of the dip and the right side of the dip. What else do you want?
BA
BA 2022년 8월 1일
Aperture that I need to calculate represents the distance between the tongue and the pharyngeal wall as attached in the image

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