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

27 visualizzazioni (ultimi 30 giorni)
JovanS il 7 Ott 2022
Commentato: JovanS il 12 Ott 2022
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 .
##### 1 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
JovanS il 11 Ott 2022

Accedi per commentare.

### Risposta accettata

Image Analyst il 11 Ott 2022
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.
##### 7 CommentiMostra 5 commenti meno recentiNascondi 5 commenti meno recenti
Image Analyst il 12 Ott 2022
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
% 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 il 12 Ott 2022
Modificato: JovanS il 12 Ott 2022
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

Accedi per commentare.

### Più risposte (1)

Benjamin Thompson il 11 Ott 2022
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.
##### 4 CommentiMostra 2 commenti meno recentiNascondi 2 commenti meno recenti
Image Analyst il 12 Ott 2022
OK, try the attached.
Snippet:
figure
% Get the histogram of classes within the mask.
edges = 0.5:1:6.5
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 il 12 Ott 2022
I attach you the final code, if you could check it, thank you very much for your help!@Image Analyst

Accedi per commentare.

### Categorie

Scopri di più su Image Segmentation and Analysis in Help Center e File Exchange

### Community Treasure Hunt

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

Start Hunting!

Translated by