Any suggestions to calculate the threshold of pixel intensity (the dip represents low intensity)
1 visualización (últimos 30 días)
Mostrar comentarios más antiguos
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');
2 comentarios
Image Analyst
el 1 de Ag. de 2022
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?
Respuesta aceptada
Image Analyst
el 1 de Ag. de 2022
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:
3 comentarios
Image Analyst
el 1 de Ag. de 2022
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?
Más respuestas (0)
Ver también
Categorías
Más información sobre Image Processing Toolbox en Help Center y File Exchange.
Productos
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!