Separating YCbCr components of an image

I am trying to seperate out the Y, Cb, and Cr components of an image and display them. I wanted the results to look like: Wiki Article Figure

I haven't quite got it right and have been trying to fix it for ages, I'm sure it's something really stupid.

    %Clear command window.
clc;
    %Clear workspace.
clear;

    %Load image 'Airplane'.
RGB = imread('Airplane.bmp'); 
    %Convert RGB image to YCbCr Components.
YCbCr = rgb2ycbcr(RGB);

    %Create a matrix of 0s, 512x512.
a = zeros(512,512);

    %Isolate Y. 
Y = YCbCr(:,:,1);
    %Isolate Cb. 
Cb = YCbCr(:,:,2);
    %Isolate Cr. 
Cr= YCbCr(:,:,3);

    %Create a YCbCr image with only the Y component.  
just_Y = cat(3, Y, a, a);
    %Create a YCbCr image with only the Cb component.  
just_Cb = cat(3, a, Cb, a);
    %Create a YCbCr image with only the Cr component.  
just_Cr = cat(3, a, a, Cr);

    %turn back to rgb
YY = ycbcr2rgb(just_Y);
CbCb = ycbcr2rgb(just_Cb);
CrCr = ycbcr2rgb(just_Cr);

    %Display the Original Image.  
figure, imshow(RGB), title('Original Image')

    %Display the Y(black and white) Component.
figure, imshow(YY), title('Y Component')
    %Display the Cb Component.
figure, imshow(CbCb), title('Cb Component')
     %Display the Cr Component.
figure, imshow(CrCr), title('Cr Component')

[Raksha Gopal Kulkarni]

Hey!!! I have a easy Mat lab code to display YCbCr components separately!!, it is consistent with the code to get RGB components from a colored image. Code:

if ~isunix()
    filename = 'path to your image';
else
    filename = 'flamingos.jpg';
end

image = imread(filename);
imshow(image);

ycbcr = rgb2ycbcr(image);
y = ycbcr(:,:,1); % y channel
cb = ycbcr(:,:,2); % cb channel
cr = ycbcr(:,:,3); % cr channel
figure();
imshow(y);

figure();
imshow(cb);

figure();
imshow(cr);

Answer 1

[Alexander De-Ville]

answered my own question. the 'a' matrix should not be 0s but 128s (50%)

[Sanik]

what does this mean ??

[Aspira Tripathy]

This means that the values concatenated along with the Y-component should have intensity of 50% and therefore, the values shouldn't be zero, but 128. This can be done by adding 128 to the zeros' matrix i.e. a = 128+zeros(512,512). Hope this helps.

Answer 2

[Dominique Sutter]

Hello, i just work on the same problem. I try the code above but it doesn't lead to good results. I've done the following : Title, and comments are in french, sorry.

clear variables
close all
clc

% Fichier image à traiter

fic='Capture.bmp';

% Lecture et Affichage de l'image
img=imread(fic);
ti=size(img);

figure(1)
imshow(img);
title(['Image originale : ',fic])
xlabel(['Résolution de l''image : ',num2str(ti(2)),'x',num2str(ti(1))])

R=img(:,:,1);
V=img(:,:,2);
B=img(:,:,3);

zer=zeros(ti,'uint8');

% Calcul

img_R=zer;
img_V=zer;
img_B=zer;

img_Y=zer;
img_Cb=zer;
img_Cr=zer;

% Séparation et affichage des canaux

img_R(:,:,1)=R;
img_V(:,:,2)=V;
img_B(:,:,3)=B;

% Calcul des Composantes
% Y (luminance - Image en niveaux de gris)
% Cb (chrominance bleue)
% Cr (chrominance rouge)

Y = 0.299*R+0.587*V+0.114*B;
Cb= -0.1687*R-0.3313*V+0.5*B+128;
Cr = 0.5*R-0.4187*V-0.0813*B+128;

figure(2)
subplot(131)
imshow(img_R)
title('Composante R')
subplot(132)
imshow(img_V)
title('Composante V')
subplot(133)
imshow(img_B)
title('Composante B')

img_Y(:,:,1)=Y;
img_Y(:,:,2)=Y;
img_Y(:,:,3)=Y;

figure(3)
imshow(img_Y);
title('Image en niveaux de gris (Luminance Y)')

img_Cb(:,:,1)=Y;
img_Cb(:,:,2)=Y;
img_Cb(:,:,3)=Y;

% Y, Cr NULL
img_Cb(:,:,1)=1.402*(-128);                         %R
img_Cb(:,:,2)=-0.34414*(Cb-128)-0.71414*(-128);     %V
img_Cb(:,:,3)=1.772*(Cb-128);                       %B

figure(4)
imshow(img_Cb);
title('Chrominance bleue (Cb)')

% Y, Cb NULL
img_Cr(:,:,1)=1.402*(Cr-128);                        %R
img_Cr(:,:,2)=-0.34414*(-128)-0.71414*(Cr-128);      %V
img_Cr(:,:,3)=1.772*(-128);                          %B

figure(5)
imshow(img_Cr);
title('Chrominance rouge (Cr)')

Answer 3

[DGM]

See also:

https://www.mathworks.com/matlabcentral/answers/580287-how-to-display-y-cb-and-cr-components-of-an-image#answer_955405

This answer explains the limits of the given method of representation, and it includes links to similar examples, including ones recreating the Wiki barn photoset.