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Can anyone please explain the working of this code ?

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Sarmad Paracha
Sarmad Paracha 2019-3-1
评论： Mark Sherstan 2019-3-4
Hello,
In this code a rule is being applied on the image to see if Ydash (luminance of the image) is greater then Cb(chrominance of blue component) but i am unable to understand the working of this code. rchannel, gchannel and bchannel are the R G B components of the image defined erlier in the code(not mentioned here) Can anyone please help me? Thanks
[R1r, R1c] = find(Ydash>Cb);
ruleIpixel=size(R1r);
Ir1= zeros(size(Im1),'uint8');
for i=1:ruleIpixel-1
Ir1(R1r(i),R1c(i),1) =rchannel(R1r(i),R1c(i));
Ir1(R1r(i),R1c(i),2) =gchannel(R1r(i),R1c(i));
Ir1(R1r(i),R1c(i),3) =bchannel(R1r(i),R1c(i));
i=i+1;
end

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采纳的回答

Mark Sherstan
Mark Sherstan 2019-3-1
Let us know if you need further clarification at any of the steps.
[R1r, R1c] = find(Ydash>Cb); % Find the indices where Ydash is greater than Cb
ruleIpixel=size(R1r); % Get the number of instances Ydash>Cb
Ir1= zeros(size(Im1),'uint8'); % Create a matrix nxn of type uint8 which is the size of some variable Im1
for i=1:ruleIpixel-1 % Create a loop incrementing by 1 for the number of instances (Ydash>Cb - 1). Note i should be replaced with ii as i is an imaginary number variable in MATLAB.
% In the zero matrix replace the zero with elements from each color channel based on the indices found on line 1 of code.
Ir1(R1r(i),R1c(i),1) =rchannel(R1r(i),R1c(i)); % For the red dimension (1)
Ir1(R1r(i),R1c(i),2) =gchannel(R1r(i),R1c(i)); % For the green dimension (2)
Ir1(R1r(i),R1c(i),3) =bchannel(R1r(i),R1c(i)); % For the blue dimension (3)
i=i+1; % Not required
end
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Mark Sherstan
Mark Sherstan 2019-3-4
Lets say that your [R1r, R1c] values are as such: (1,3) and (2,2) and (3,2) and that your zero matrix is as so:
0 0 0
0 0 0
0 0 0
We know that an RGB image can be simplified into the following:
Capture.PNG
So the loop would give a result as so:
Iteration 1:
0 0 121 0 0 153 0 0 49
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
Iteration 2:
0 0 121 0 0 153 0 0 49
0 195 0 0 205 0 0 34 0
0 0 0 0 0 0 0 0 0
Iteration 3: Although there are 3 entries found using find the for statment considers n-1 of them so the last one is not conisdered.

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