Average histogram of R,G,B histograms

I need to find the avrege of histograms B,R,G that I already found using this code so can you please help me?

after finding the average I need to use it as the basis to obtain a single histogram equalization intensity transformation function so how is that specifically done?

this the code I used

X = im2double(imread('Capture.PNG'));
R = X(:,:,1);
G = X(:,:,2);
B = X(:,:,3);
z = zeros(size(R));
Rimg = cat(3, R, z, z);
Gimg = cat(3, z, G, z);
Bimg = cat(3, z, z, B);
L256 = linspace(0,1,256).';
z256 = zeros(256,1);
mapR = [L256, z256, z256];
mapG = [z256, L256, z256];
mapB = [z256, z256, L256];
figure; image(Rimg); colormap(mapR); colorbar();
figure; image(Gimg); colormap(mapG); colorbar();
figure; image(Bimg); colormap(mapB); colorbar();


I=imread('Capture.PNG');
counts1=imhist(I(:,:,1));
counts2=imhist(I(:,:,2));
counts3=imhist(I(:,:,3));
figure, plot(counts1,'r')
figure, plot(counts2,'g')
figure, plot(counts3,'b')

mean_counts = mean([counts1(:), counts2(:), counts3(:)], 2);
figure, plot(mean_counts)

help me please

Answer 1

[Ridwan Alam]

I=imread('Capture.PNG');

[counts1,edges1]=histcounts(I(:,:,1),linspace(0,1,256));
[counts2,edges2]=histcounts(I(:,:,2),linspace(0,1,256));
[counts3,edges3]=histcounts(I(:,:,3),linspace(0,1,256));

figure, plot(edges1(1:end-1),counts1,'r')
figure, plot(edges2(1:end-1),counts2,'g')
figure, plot(edges3(1:end-1),counts3,'b')

mean_counts = mean([counts1(:), counts2(:), counts3(:)], 2);
figure, plot(edges1(1:end-1),mean_counts)

J = histeq(I,mean_counts)

[Ridwan Alam]

Sorry ... i meant histcounts()

[Image Analyst]

So what is the length of edges1 and counts1?

[Maryam Al-Muhaini]

X = im2double(imread('Capture.jpg'));
R = X(:,:,1);
G = X(:,:,2);
B = X(:,:,3);
z = zeros(size(R));
Rimg = cat(3, R, z, z);
Gimg = cat(3, z, G, z);
Bimg = cat(3, z, z, B);
L256 = linspace(0,1,256).';
z256 = zeros(256,1);
mapR = [L256, z256, z256];
mapG = [z256, L256, z256];
mapB = [z256, z256, L256];
figure; image(Rimg); colormap(mapR); colorbar();
figure; image(Gimg); colormap(mapG); colorbar();
figure; image(Bimg); colormap(mapB); colorbar();

I=imread('Capture.jpg');
counts1=imhist(I(:,:,1));
counts2=imhist(I(:,:,2));
counts3=imhist(I(:,:,3));
figure, plot(counts1,'r')
figure, plot(counts2,'g')
figure, plot(counts3,'b')

I=imread('Capture.jpg');
counts1;edges1=histcounts(I(:,:,1),linspace(0,1,256));
[countedges2]=histcounts(I(:,:,2),linspace(0,1,256));
[counts3,edges3]=histcounts(I(:,:,3),linspace(0,1,256));
figure, plot(edges1(1:end-1),counts1,'r')
figure, plot(edges2(1:end-1),counts2,'g')
figure, plot(edges3(1:end-1),counts3,'b')
mean_counts = mean([counts1(:), counts2(:), counts3(:)], 2);
figure, plot(edges1(1:end-1),mean_counts)
J = histeq(I,mean_counts);

This is the code

[Ridwan Alam]

So ... is it working now? what did you mean by the results being incovenient?

[Maryam Al-Muhaini]

The code works, but I get a line for histogram + I believe we’re asked to show an image at the end after we do all that bit these so changes

[Ridwan Alam]

Figure; imshow(J)

[Maryam Al-Muhaini]

it's suppoe to enhance the color of the picture but i don't think that this is enhanced.th graphs im getting are weird as it's shown. please check the question i sent

Histogram-equalize the R, G, and B images separately using the histogram-equalization program and convert the image back to tif format.(b) Form an average histogram from the three histograms in(a) and use it as the basis to obtain a single histogram equalization intensity transformation function. Apply this functionto the R, G, and B components individually, and convert the results to jpg. Compare and explain the differences in the tif images in (a) and (b).

i have to submit it by tomorrow please help

[Image Analyst]

Sorry - I guess I should have told you that this code does not do what you asked or what the question asked. Not sure why you accepted this answer. But since I'm seeing it after you've already turned it in, it's too late now to fix it. If you're going to use the built-in histeq() and you pass in the mean histogram of all the channels, then it will produce an image that attempts to match that mean histogram. That's not what was asked. Plus the first argument ot histeq() must be a gray scale image, not a color image. See the help:

"J = histeq(I,hgram) transforms the grayscale image I so that the histogram of the output grayscale image J with length(hgram) bins approximately matches the target histogram hgram."

Too bad. Oh well, like I said, too late. Should have taken my advice below.

[Ridwan Alam]

Oh. So sorry just saw these comments, specially the original question.

X = im2double(imread('peppers.PNG'));
R = X(:,:,1);
G = X(:,:,2);
B = X(:,:,3);

(a) Histogram-equalize the R, G, and B images separately using the histogram-equalization program and convert the image back to tif format.

Req = histeq(R,256); figure; imshow(Req);
Geq = histeq(G,256); figure; imshow(Geq);
Beq = histeq(B,256); figure; imshow(Beq);
Xeq = cat(3,Req,Geq,Beq); 
figure;imshow(Xeq); 
% it looks different than X because of the separate hist equalization of R,G,B

(b) Form an average histogram from the three histograms in(a) and use it as the basis to obtain a single histogram equalization intensity transformation function. Apply this functionto the R, G, and B components individually, and convert the results to jpg. Compare and explain the differences in the tif images in (a) and (b).

[countsR,edgesR]=histcounts(R,linspace(0,1,256));
[countsG,edgesG]=histcounts(G,linspace(0,1,256));
[countsB,edgesB]=histcounts(B,linspace(0,1,256));
figure, plot(edgesR(1:end-1),countsR,'r')
figure, plot(edgesG(1:end-1),countsG,'g')
figure, plot(edgesB(1:end-1),countsB,'b')

mean_counts = mean([countsR(:), countsG(:), countsB(:)], 2);
figure, plot(edgesR(1:end-1),mean_counts)

Reqm = histeq(R,mean_counts);
figure;imshow(Reqm)
Geqm = histeq(G,mean_counts);
figure;imshow(Geqm)
Beqm = histeq(B,mean_counts);
figure;imshow(Beqm)
% convert 
Xeqm = cat(3,Reqm,Geqm,Beqm);
figure;imshow(Xeqm)

[Image Analyst]

No, I don't think so. Again, your code will just do histogram matching, like I and the help said. You can't pass in some histogram with some non-flat shape if you want a flat output histogram. You can do that for a single gray scale image, but for the color image task it asked, you have to take the mean_counts, then compute the CDF with cumsum() then invert it and use intlut() to transform each color channel and then use cat() to recombine them. You can't do what it asked with the built-in histeq() function.

Answer 2

[Image Analyst]

I wouldn't do it like that. You'd probably have the least annoying result if you transformed to HSV color space with rgb2hsv(), then called histeq() on the V channel ONLY, then get back to RGB space with hsv2rgb(). That said, usually histogram equalization looks crummy - it's just a mathematical curiosity that beginners like to try but no one really uses to make pleasing pictures. Once you try it you'll see what I mean. Bad pictures. Would be better to just use a gamma or linear stretch rather than the unnatural stretch of histogram equalization which gives harsh, lousy looking images.

[Maryam Al-Muhaini]

unfortunatlly, I'm doing a homework and I've been asked to do this :(

[Image Analyst]

Then you can still call rgb2gray() then get the histogram, then the CDF by using cumsum(), and then the transform. Then apply it to each color channel.