Passing a matrix to graycomatrix

Could you please help me with the following question?
My code reads a DICOM image. The image is 512 * 512 uint 16.
The code draws an ROI on the image. After creating a mask, the code gets X and Y coordinates as well as the intensity of each pixel within the mask.
The result is a matrix of n by m, which its value corresponds to the image intensity. I want to know how I can pass this matrix to the graycomatrix function?

 Respuesta aceptada

Image Analyst
Image Analyst el 29 de Sept. de 2021

0 votos

See attached example.

3 comentarios

Thank you Image Analyst. But, I could not find the answer to my question. I want to know how I should pass the matrix to the graycomatrix function.
I use the following code to create the matrix:
for i = 1:size(n)
iy = XROI(n(i, 1));
ix = YROI(n(i, 1));
intens = PixelValues(n(i, 1));
imgmap(ix, iy) = intens;
end
Where n represents the number of the pixels within the mask. XROI is the X-coordinates of all pixels within the mask. YROI is the Y-coordinates of all pixels within the mask. intens is intensity of the corresponding pixel.
The imgmap is the matrix
Then, should I pass the matrix as an image to the graycomatrix function?
If yes, which of the following code is correct:
NewImage = uint8(imgmap);
or given the original image is uint 16, I shoudl use:
NewImage = uint16(imgmap);
glcm = graycomatrix(NewImage, 'Offset', offsets);
or I can directly plugin the matrix:
glcm = graycomatrix(imgmap, 'Offset', offsets);
Walter Roberson
Walter Roberson el 30 de Sept. de 2021
Editada: Walter Roberson el 30 de Sept. de 2021
imgmap = zeros(max(YROI(n(:,1))), max(XROI(n(:,1))), 'like', PixelValues);
for i = 1:size(n, 1)
iy = XROI(n(i, 1));
ix = YROI(n(i, 1));
intens = PixelValues(n(i, 1));
imgmap(iy, ix) = intens;
end
glcm = graycomatrix(imgmap, 'Offset', offsets);
Note: you had x and y reversed. X is horizontal distance, which is column number. y is vertical distance, which is row number.
Nathan Armani
Nathan Armani el 30 de Sept. de 2021
Thank you

Iniciar sesión para comentar.

Más respuestas (0)

Preguntada:

el 29 de Sept. de 2021

Comentada:

el 30 de Sept. de 2021

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by