How to change the color of each pixel in a masked image?

I found the Euclidean distances between each pixel of the mask and six reference colors. Now i want to change the color of every pixel which is in the skin lesion (masked image). Each pixel has to change color and will take the color of the shortest euclidean distance ( for example if the shortest euclidean distance is the one with red color , the pixel will turn into red). So I have to create a new image in which every pixel of mask will obtain one color of the 6 reference colors. Then the last step is to check if the number of pixels belonging to each reference colour exceeds by 5% of the total number of pixels of the lesion. If this occurs the color score is incremented by 1, otherwise the color score of each color is 0.

Any idea for doing this procedure ?

I attach my matlab code so far .

[JovanS]

@Image Analyst could you please help me ?

Answer 1

[Image Analyst]

You just didn't use the min function like I suggested in your other post. Here's the relevant snippet. The full demo is attached.

% Stack all the delta E into a 3-D image.
stack_array = cat(3,deImage_bg,deImage_wh,deImage_bl,deImage_red,deImage_db,deImage_lbr);
% Produce a classified image where the value is the slice that has the lowest delta E.
[~, classifiedImage] = min(stack_array, [], 3);
hFigClassified = figure;
imshow(classifiedImage, [])
impixelinfo;
cmap = lines(6);
colormap(cmap);
hcb = colorbar(gca,'Ticks', 1:6, 'TickLabels', {'BlueGray', 'White', 'Black', 'Red', 'Dark Brown', 'LightBrown'});
hcb.Label.String = 'Color Category';
hold on;
visboundaries(lesionBoundary);
hold off;
title('Classified Image', 'FontSize',fontSize)
drawnow;
hFigClassified.WindowState = "maximized";

Here is the image showing the classifications. As you can see, most of the melanoma is classified as being in the dark brown category.

[JovanS]

Thank you very much this is exactly what i wanted to do , one more last question. how can i check if the number of pixels belonging to each reference colour exceeds by 5% of the total number of pixels of the lesion? @Image Analyst

[Image Analyst]

To get the fraction of each class within the masked lesion area, simply take the histogram of the classified image:

figure
% Get the histogram of classes within the mask.
histObj = histogram(classifiedImage(mask), 'Normalization','probability')
% Show fractions of each class in the mask area.
probs = histObj.Values

probs =

Columns 1 through 4

0.000903117785839316 0 0.00423991814438046 0

Column 5

0.99485696406978

[JovanS]

I can't understand how can the histogram could be helpful for me. i have to found the percentage of existance of each color in the skin lesion(to be clear the percentage for all reference colors) . How can histogram help to achive this goal? @Image Analyst

[Image Analyst]

Not sure why you don't understand. The classified image is the color class that has the minimum delta E from the image pixel to the 6 reference colors. So if a particular pixel has a delta E of 2 for color #6 and a delta E of more than 10 for all the other colors, then that pixel is classified as being of color #6. The classified image has only values of 1-6 for the 6 colors. So if you take a histogram, inside the lesion area, it will count how many pixels have value 1, how many pixels have value 2, and so on. So you get a count of how many pixels are in each color class. Divide that by the number of pixels in the lesion mask region and you turn the count into a percentage. So what's not to understand?

[JovanS]

First of all , this line of code:

% Show fractions of each class in the mask area.

probs = histObj.Values

It give us as an exit of five values(why not for 6 colors?), and the values are numbers in a scientific notation and not into decimal notation. So i dont understand how to utilize these values in order to find the percentage of each color. @Image Analyst

[Image Analyst]

One of the colors (#1) was not in the mask area so the histogram did not include that one. It's probably best to specify the edges for histogram so that the first and last bin are known and constant.

figure
% Get the histogram of classes within the mask.
edges = 0:6
histObj = histogram(classifiedImage(mask), edges, 'Normalization','probability')
% Show fractions of each class in the mask area.
probs = histObj.Values

probs =

Columns 1 through 4

0 0.000903117785839316 0 0.00423991814438046

Columns 5 through 6

0 0.99485696406978

[JovanS]

ok thank you but one last question , column 1corresponds to bluegray , 2 to white , 3 to black , 4 to red , 5 to darkbrown and 6 to lightbrown?? because according to this , the most number of pixels are lightbrown. As we saw in classifiedimage the melanoma is classified as being in the dark brown category. @Image Analyst

Answer 2

[Benjamin Thompson]

See if this helps. There is probably a faster way to do it with color mapping functions in the Image Processing Toolbox but it does not take very long on your image size.

[JovanS]

Sorry but I can't understamd how this answer could help me. To be clear , i have already found the Euclidean distance between each pixel of lesion's mask and each reference color. Now i dont know how to find the smallest distance between each pixel and references colors in order to label for example a piel as a red pixel , another one as a lightbrown pixel etc according to its euclidean distance. Could you please help me? @Benjamin Thompson

[Image Analyst]

Let me look into it. If you mouse around over the image you'll see the majority of pixels are 5 as you can see in the lower left corner of the figure. Let me try to figure out why the histogram is not showing that.

[Image Analyst]

OK, try the attached.

Snippet:

figure
% Get the histogram of classes within the mask.
edges = 0.5:1:6.5
pixelValuesInMaskOnly = classifiedImage(mask);
histObj = histogram(pixelValuesInMaskOnly, edges, 'Normalization','probability')
grid on;
ylabel('Area Fraction (0 - 1)')
xticklabels(colorNames);
title('Fractions for Each Color Class')
% Show fractions of each class in the mask area.
probs = histObj.Values

probs =

Columns 1 through 4

0.000903117785839316 0 0.00423991814438046 0

Columns 5 through 6

0.99485696406978 0

[JovanS]

I attach you the final code, if you could check it, thank you very much for your help!@Image Analyst