MATLAB: Gray Matter Area and Volume Calculations

binaryimage processingImage Processing Toolboximage segmentationMATLAB

Hello, I am working on gray matter area and volume calculation of brain MRI image. I have some doubts with my code below:
% Area Calculation
Pixel_Numbers = sum(J(:));
DistanceinUnit = 1.796875; %Based on pixel spacing (in mm?)
DistanceinPixel = sqrt(x^2+y^2);
DistancePerPixel = DistanceinUnit/DistanceinPixel;
Area = (Pixel_Numbers)*(DistancePerPixel^2); %Is it in mm^2 or mm^2/pixel?
% Volume Calculation (Is the formula correct?)
LengthinPixel = length(find(J(:)==1));
Length = LengthinPixel*DistanceinUnit;
Volume = Area*Length;
The information about pixel spacing is 1.796875\1.796875. I'd be glad if you can help me to figure out the answer. Here is the picture. Thank you.

Best Answer

Hello, i think you're image appears to be binarized already. This means we can obtain the area well with the regionprops function. https://www.mathworks.com/help/images/ref/regionprops.html.
In terms of the volume, are you wanting to sum up the number of pixel in that area * intensity? In you case if the image is binary, the volume and area will be the same. volume=area*height but height is 1 because your image is binary.
The last thing you want now is pixels to mm? just determine the conversion for this, if a pixel is 1.796875x1.796875 we will have an area=(1.796875)^2*num_pixels.
If all the area is one block, it will be returned from regionprops, if not you'll need to sum it up. Here is my logic for this problem:
image=imread('J.jpg');
image=rescale(image,0,1); %helps for binarization
BW_im=imbinarize(image); %see documentation, it is versatile
hh=regionprops(BW_im,{'Centroid','Area',}); %centroid for locations.
total_area=0;
total_volume=0;
height=1;%your image is binary, if it's not this will change pixel by pixel.
%see the regionprops function to sum up pixel values as well.
for i=1:length(hh)
total_area=total_area+hh(i).Area;
total_volume=total_volume+(hh(i).Area*height)
end
area_in_mm=(1.796875.^2)*total_area;
Hope this helps, if not, drop a comment.
RC