Sorting unique colors from the image (much simplified now)

Hello everyone,

Code uses rgb2hex and hex2rgb from FEX, I might be doing unecessary work with this, but let it be for now - https://ch.mathworks.com/matlabcentral/fileexchange/46289-rgb2hex-and-hex2rgb.

clc;
clear;
fclose all;
File_Name = 'https://upload.wikimedia.org/wikipedia/commons/thumb/e/ec/Mona_Lisa%2C_by_Leonardo_da_Vinci%2C_from_C2RMF_retouched.jpg/161px-Mona_Lisa%2C_by_Leonardo_da_Vinci%2C_from_C2RMF_retouched.jpg'; % Image name
Image_Data = imread(File_Name);

[V_Pixels, H_Pixels, ~] = size(Image_Data);
Num_Points = H_Pixels * V_Pixels; 
Colors = cell(V_Pixels, H_Pixels);

% Getting colors as hex values
for ii = 1 : 1 : V_Pixels
    for jj = 1 : 1 : H_Pixels
        Colors {ii, jj} = rgb2hex([Image_Data(ii,jj,1) Image_Data(ii,jj,2) Image_Data(ii,jj,3)]); % RGB values
    end
end

Color_Vector = (unique(reshape(Colors, [1, Num_Points])))'; %Removes duplicates and sorts
Color_RGB = hex2rgb(Color_Vector);


x = 1:1:length(Color_Vector);
y = ones(1,length(Color_Vector));

figure(1);
hb = bar(x, y);
hb.FaceColor = 'flat';
for ii = 1 : 1 : length(Color_Vector)
    hb.CData(ii,:) = Color_RGB(ii,:);
end

A graph is attached here, the colors are not sorted well. Obviously it means that colors can't be sorted this way.

I found some posts on sorting the image by HSV, but also realised that I am out of depth when it comes to image processing. I found an example here https://stackoverflow.com/a/2246189/14022732, but I am a bit lost when I have to convert back to rgb.

If you could share some tips on this, it would be great.

EDIT: I would like to sort unique colors so that they appear like a gradient, see my reply below.

Answer 1

[Image Analyst]

Mario:

Here is the color gamut of that image in RGB color space:

and here is the gamut in HSV color space:

Each dot in the conical gamut represents the color of one pixel in the image. You can see that there is no real sensible way to sort that. What you're doing has nothing to do with HSV color space. You're all in RGB color space and you're sorting by Blue first, then green, and finally by red. Not sure that makes any sense as far as sorting goes, and I don't see how sorting by anything in HSV color space makes sense either. Do you just want to see unique colors?

Also I don't know what you mean by converting back to RGB. Not sure exactly what would get converted back - all the values, or just the unique values. If you keep all the values, of course you can convert back but if you throw out some because they're duplicated, then you can convert back to colors but you won't have an image, list a list of RGB values because you threw out some pixels and we don't know what to put there. For example if we have an aray m=[1,2,3,3,4] and we multiply by 2 we get m2 = [2,4,6,6,8]. Now if we throw out duplicates we get m2 = [2,4,6,8]. So you can divide by 2 to get [1,2,3,4] but that is not what you originally started with.

[Mario Malic]

Hi Image Analyst,

Thank you very much for your reply.

Sorry for missing out on precisely saying what I wanted to do in the post. As you can see on Image Inspector, the picture has 23007 unique colors. I would like to put these 23007 colors in a vector and in "order" so that they would appear like a gradient on picture below (just an example, not obtained from image).

[Stephen23]

"...so that they would appear like a gradient..."

But they aren't arranged along any kind of simple "gradient". They are some (essentially random) colors represented by points in a 3D colorspace, for which there is no single concept of "sorting" or "order": the internet already has a lot of discussions on this topic.

Another way to think about this: your example image appears to show one hue over a range of chroma: where should other hues be represented on that "gradient" example? For example, the blue water, where should that color appear in your example?

[Mario Malic]

Thank you very much for the comment Steven,

Alright, so, there's not a simple way to sort the colors.

If there's only one hue on my image, then the blue water can't be represented, since it is another hue value?

To add up on this question, I am only interested in RGB values since I can only use them as an input for other program. An alternative solution is that I would find min and max RGB values and would create a vector of integer values between their range and that seems to be okayish solution as well.

[Image Analyst]

You need to come up with some sort of metric to sort on. Perhaps you just want to use the distance from (0,0,0). But then (128, 0, 0) and (0, 128, 0) and (0, 0, 128) would all have the same vector length from (0,0,0) but they're all drastically different colors. So which would you put first? Maybe sort all things that distance from the origin by hue. But then you have another problem: even with the same distance and same hue, colors could have different saturation so you'd have to sort by that too. So you end up having to sort first by one thing, then by another, then by a third. A linear scale like you have would thus show many "ramps" where that section has the same first sorting metric. Maybe that's okay, maybe not. If not, maybe you sort by two and collapse the third dimension into a plane, kind of like if you looked straight down the V axis:

If that's still not OK, then you're back to plotting the full gamut in 3-D. You can use colorcloud() for that.

[Mario Malic]

I really thank you both for your efforts, kinda helped me to realize this is not something easy to do or something that someone should do without proper knowledge. I could spend days trying to figure out how should I sort these values and get to nowhere.

So, if you agree, I think it's the best to close/delete this topic.

[Image Analyst]

Actually I think it's a good discussion that others may learn from. I'm not saying it's a worthless exercise, in fact I think you should try some things. I'm just saying that you should not expect, at least for general images, that you'll get a nice smooth ramp/gradient. So try it out and see what it looks like. It could be fun. Why not just start with the distance of all colors from black? Then if there are multiple pixels with that same distance, just take them as they are or else pick them randomly. The output image might look interesting. Here's a start.

% Demo to find distances.  By Image Analyst, May 5, 2020.
clc;    % Clear the command window.
close all;  % Close all figures (except those of imtool.)
clearvars;
workspace;  % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 16;
fprintf('Beginning to run %s.m ...\n', mfilename);

%-----------------------------------------------------------------------------------------------------------------------------------
% Read in image.
folder = [];
baseFileName = 'peppers.png';
fullFileName = fullfile(folder, baseFileName);
% Check if file exists.
if ~exist(fullFileName, 'file')
	% The file doesn't exist -- didn't find it there in that folder.
	% Check the entire search path (other folders) for the file by stripping off the folder.
	fullFileNameOnSearchPath = baseFileName; % No path this time.
	if ~exist(fullFileNameOnSearchPath, 'file')
		% Still didn't find it.  Alert user.
		errorMessage = sprintf('Error: %s does not exist in the search path folders.', fullFileName);
		uiwait(warndlg(errorMessage));
		return;
	end
end
rgbImage = imread(fullFileName);
[rows, columns, numberOfColorChannels] = size(rgbImage)
% Display the test image full size.
subplot(2, 1, 1);
imshow(rgbImage, []);
axis('on', 'image');
caption = sprintf('Reference Image : "%s"', baseFileName);
title(caption, 'FontSize', fontSize, 'Interpreter', 'None');
drawnow;
hp = impixelinfo(); % Set up status line to see values when you mouse over the image.


% Set up figure properties:
% Enlarge figure to full screen.
hFig1 = gcf;
hFig1.Units = 'Normalized';
hFig1.WindowState = 'maximized';
% Get rid of tool bar and pulldown menus that are along top of figure.
% set(gcf, 'Toolbar', 'none', 'Menu', 'none');
% Give a name to the title bar.
hFig1.Name = 'Demo by Image Analyst';

[r, g, b] = imsplit(double(rgbImage));
distances = sqrt(r(:).^2 + g(:).^2 + b(:).^2);
[sortedDistances, sortIndexes] = sort(distances, 'ascend');
% Sort our RGB arrays the same way.
r = r(sortIndexes);
g = g(sortIndexes);
b = b(sortIndexes);
% Now determine the number of rows and columns that would make a good output size.
% TODO : you do this.  You might use factors or some function to determine two
% integers that are the biggest.  You can use the original image's rows and columns if you want:
[rows, columns, numberOfColorChannels] = size(rgbImage);
% Then use reshape to make r, g, and b into a rectangular matrix of that size.
reshapedR = reshape(r, rows, columns);
reshapedG = reshape(g, rows, columns);
reshapedB = reshape(b, rows, columns);
% Then create the final rgb image like
rgbOut = cat(3, uint8(reshapedR), uint8(reshapedG), uint8(reshapedB));
subplot(2, 1, 2);
imshow(rgbOut);
title('Sorted Colors', 'FontSize', fontSize, 'Interpreter', 'None');
drawnow;
hp = impixelinfo(); % Set up status line to see values when you mouse over the image.

You can see what I mean about it not being smooth and monotonically increasing - that's why there's "noise" in the output image. You can reduce that be doing a further sort of all distances that are the ""same" or close to it.

[Mario Malic]

Sorry for the delay, I did a bit of the work that you suggested to me and it was much better than what I had done. Even though images had more colors, I had issues with yellowish ones as barely appeared. I have spent some time trying to fix it, and then gave up on this to try to solve my problem the other way.

Which was the right way from the start, I just haven't seen it. I just had to reshape the Image data into a vector, get colors in hex format and put them back into other program in the same order.

Thank you very much, both, for your comments as I learned a little bit more about images and the neat stuff reshape can do.

Image tax, you can see the order of the colors on the right, it is just the same order as they appear on the image when one would reshape by the rows. and the band is upside down

[Image Analyst]

There is no "right" way to do it. There are several ways to do it, according to the respective algorithms, but it's not like one is more correct than another - they're just different.

[Mario Malic]

I meant, the right way to solve my problem, not the right way to sort the colors.