# use rgb to caculate three brown ratio

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bozheng il 6 Apr 2024
Risposto: Image Analyst il 8 Apr 2024
sorry i say more about i want coding which can show light color, dark color, intermediate color ratio
focus is kind and ratio
image_double = im2double(image);
percentage_a = image_double(:,:,1) / sum(image_double, 3);
percentage_b = image_double(:,:,2) / sum(image_double, 3);
percentage_c = image_double(:,:,3) / sum(image_double, 3);
disp([num2str(percentage_a) '%']);
disp([num2str(percentage_b) '%']);
disp([num2str(percentage_c) '%']);
and picture like it ( the word we can ignore)
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### Risposta accettata

DGM il 6 Apr 2024
Modificato: DGM il 6 Apr 2024
The question is unclear. This seems to be what was attempted, but it doesn't make sense given the question statement.
inpict = im2double(inpict);
% you can take the ratio of RGB channels versus the sum
% but this tells us nothing about the population of brown pixels
% or the three colors themselves
channelratio = mean(inpict./sum(inpict,3),1:2);
channelratio = permute(channelratio,[1 3 2])
channelratio = 1x3
0.5123 0.3567 0.1310
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Instead, I think it makes more sense to find the relative area covered by each of the three brown colors.
% instead of doing this using explicit color matching
% i'm just going to use rgb2ind().
inpict = im2double(inpict);
ncolors = 4; % three browns and one white
[indpict CT] = rgb2ind(inpict,ncolors,'nodither');
% CT should represent three browns and one near-white.
% find the white tuple, so that we can ignore it.
[~,whidx] = max(sum(CT,2)); % the index of white
bridx = setdiff(1:ncolors,whidx); % the indices of browns
% find the pixels belonging to each brown color
nbrownpix = zeros(1,3);
for k = bridx
nbrownpix(k) = nnz(indpict == k-1); % sum
end
nbrownpix = nbrownpix./numel(indpict) % normalize
nbrownpix = 1x3
0.2576 0.3449 0.3334
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sum(nbrownpix) % total coverage of brown
ans = 0.9359
This can always be done doing any sort of color segmentation. There are plenty of examples on the forum. I'm just going to use the same colors as before.
% we can segment the image into three masks,
% one for each shade of brown.
inpict = im2double(inpict);
CT = [0.251 0.1843 0.01569; % middle
0.8902 0.7961 0.6392; % left
0.6196 0.3255 0.01961]; % right
tol = 0.1;
ncolors = size(CT,1);
nbrownpix = zeros(1,ncolors);
for k = 1:ncolors
% reorient tuple along page axis
pickedcolor = permute(CT(k,:),[1 3 2]);
% perform box (range) selection in RGB
% sum
end
nbrownpix = 1x3
0.2439 0.3253 0.3209
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bozheng il 6 Apr 2024
i mean is it will show light color, dark color, intermediate color ratio= 30%:40%;30%(total is approch 95-100%) sorry and thank
DGM il 6 Apr 2024
Modificato: DGM il 6 Apr 2024
That's what the examples I gave do, although I express the results as unit fractions. Perhaps what's left unclear is which fraction corresponds to each of the three colors. I'll make some modifications. Will update in a bit.
EDIT:
Here are both examples, using very similar output formatting to express the results.
% instead of doing this using explicit color matching
% i'm just going to use rgb2ind().
inpict = im2double(inpict);
ncolors = 4; % three browns and one white
[indpict CT] = rgb2ind(inpict,ncolors,'nodither');
% CT should represent three browns and one near-white.
% find the white tuple, so that we can ignore it.
[~,whidx] = max(sum(CT,2)); % the index of white
bridx = setdiff(1:ncolors,whidx); % the indices of browns
% find the pixels belonging to each brown color
nbrownpix = zeros(1,3);
for k = bridx
nbrownpix(k) = nnz(indpict == k-1); % sum
end
nbrownpix = nbrownpix./numel(indpict); % normalize
% find order of shades in CT (darkest -> lightest)
[~,idx] = sort(im2gray(permute(CT(bridx,:),[1 3 2]))); % BT601 luma
ordstr = {'darkest','middle','lightest'}; % assuming three shades
for k = 1:numel(idx)
thistuplestr = mat2str(CT(idx(k),:),4);
fprintf('The %s brown is approximately %s in RGB\n', ordstr{k},thistuplestr)
fprintf('It covers %.1f%% of the total image area\n\n',nbrownpix(idx(k))*100)
end
The darkest brown is approximately [0.251 0.1843 0.01569] in RGB
It covers 25.8% of the total image area
The middle brown is approximately [0.6196 0.3255 0.01961] in RGB
It covers 33.3% of the total image area
The lightest brown is approximately [0.8902 0.7961 0.6392] in RGB
It covers 34.5% of the total image area
% we can segment the image into three masks,
% one for each shade of brown.
inpict = im2double(inpict);
CT = [0.251 0.1843 0.01569;
0.8902 0.7961 0.6392;
0.6196 0.3255 0.01961];
tol = 0.1;
ncolors = size(CT,1);
nbrownpix = zeros(1,3);
for k = 1:ncolors
% reorient tuple along page axis
pickedcolor = permute(CT(k,:),[1 3 2]);
% perform box (range) selection in RGB
% sum
end
% find order of shades in CT (darkest -> lightest)
[~,idx] = sort(im2gray(permute(CT(bridx,:),[1 3 2]))); % BT601 luma
ordstr = {'darkest','middle','lightest'}; % assuming three shades
for k = 1:numel(idx)
thistuplestr = mat2str(CT(idx(k),:),4);
fprintf('The %s brown is approximately %s in RGB\n', ordstr{k},thistuplestr)
fprintf('It covers %.1f%% of the total image area\n\n',nbrownpix(idx(k))*100)
end
The darkest brown is approximately [0.251 0.1843 0.01569] in RGB
It covers 24.4% of the total image area
The middle brown is approximately [0.6196 0.3255 0.01961] in RGB
It covers 32.1% of the total image area
The lightest brown is approximately [0.8902 0.7961 0.6392] in RGB
It covers 32.5% of the total image area
Intuition should agree with these fractions. The darkest shade is the most occluded by the text, so it has the lowest coverage. The other two shades should have similar coverage.
The results differ slightly. In the first example, the results are mutually-exclusive. In the second, they are not necessarily mutually-exclusive. The second result also relies on some prior sampling to determine the representative colors for each brown region.
FWIW, I tend to regard "percent" as purely a thing that exists in text displays. As far as I'm concerned, "one hundred percent" is written "1". Unless it has a % character appended to it, "100" means "one hundred". Text outputs are for human use, whereas numeric outputs are for programmatic use. That's why I dump unit fractions when asked for percent.

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### Più risposte (2)

Hassaan il 6 Apr 2024
image_double = im2double(image); % Convert to double precision floating point
% Calculate the sum of RGB values across the third dimension
sum_rgb = sum(image_double, 3);
% Calculate the percentage of each RGB channel relative to the sum
percentage_r = mean(mean(image_double(:,:,1) ./ sum_rgb));
percentage_g = mean(mean(image_double(:,:,2) ./ sum_rgb));
percentage_b = mean(mean(image_double(:,:,3) ./ sum_rgb));
% Display the results
fprintf('Red Channel Percentage: %.2f%%\n', percentage_r * 100);
fprintf('Green Channel Percentage: %.2f%%\n', percentage_g * 100);
fprintf('Blue Channel Percentage: %.2f%%\n', percentage_b * 100);
-----------------------------------------------------------------------------------------------------------------------------------------------------
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bozheng il 6 Apr 2024
sorry i say more about i want coding which can show light color, dark color, intermediate color ratio

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Image Analyst il 8 Apr 2024
You could try using kmeans with k=5 for the three browns, black, and white (for the letters). Or you could also use K nearest Neighbors. Demos are attached.
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