How can I get this code to show only the veins in a fundus image of an eye ?

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Abhishek Deshpande
Abhishek Deshpande 2023-3-14
回答： Image Analyst 2023-3-15
I have been trying to run a code in order to extract the retinal veins fromt he fundus image of an eye however it keeps giving me such erroneous output. I am including the part for extracting veins , however I am unsure so as to what is going wrong with this.
Code:
Coloured = imread('fundus_image');
Converted_Image = im2double(Coloured);
Lab_Image = rgb2lab(Converted_Image);
fill = cat(3, 1,0,0);
Filled_Image = bsxfun(@times, fill, Lab_Image);
Reshaped_Lab_Image = reshape(Filled_Image, [], 3);
[C, S] = pca(Reshaped_Lab_Image);
S = reshape(S, size(Lab_Image));
S = S(:, :, 1);
Gray_Image = (S-min(S(:)))./(max(S(:))-min(S(:)));
Enhanced_Image = adapthisteq(Gray_Image, 'numTiles', [8 8], 'nBins', 256);
Avg_Filter = fspecial('average', [9 9]);
Filtered_Image = imfilter(Enhanced_Image, Avg_Filter);
Substracted_Image = imsubtract(Filtered_Image,Enhanced_Image);
Image = im2uint8(Substracted_Image(:));
[Histogram_Count,Bin_Number]=imhist(Image);
i = 1;
Cumulative_Sum = cumsum(Histogram_Count);
T(i) = (sum(Bin_Number.*Histogram_Count))/Cumulative_Sum(end);
T(i)=round(T(i));
Cumulative_Sum_2 = cumsum(Histogram_Count(1:T(i)));
MBT=sum(Bin_Number(1:T(i)).*Histogram_Count(1:T(i)))/Cumulative_Sum_2(end);
Cumulative_Sum_3=cumsum(Histogram_Count(T(i):end));
MAT=sum(Bin_Number(T(i):end).*Histogram_Count(T(i):end))/Cumulative_Sum_3(end);
i = i+1;
T(i) = round((MAT+MBT)/2);
while abs(T(i)-T(i-1))>=1
Cumulative_Sum_2 = cumsum(Histogram_Count(1:T(i)));
MBT=sum(Bin_Number(1:T(i)).*Histogram_Count(1:T(i)))/Cumulative_Sum_2(end);
Cumulative_Sum_3=cumsum(Histogram_Count(T(i):end));
MAT=sum(Bin_Number(T(i):end).*Histogram_Count(T(i):end))/Cumulative_Sum_3(end);
i = i+1;
T(i) = round((MAT+MBT)/2);
Threshold = T(i);
end
level = (Threshold - 1) / (Bin_Number(end) - 1);
Binary_Image = imbinarize(Substracted_Image, level-0.008);
Clean_Image = bwareaopen(Binary_Image, 100);
figure, imshow(Clean_Image);
--> Any help is most appreciated.

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Image Analyst
Image Analyst 2023-3-15

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