how to estimate bright channel for color image in Matlab??

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ghada sandoub
ghada sandoub 2019-3-11
评论： ghada sandoub 2019-3-13
how to estimate the bright channel prior for color image (convert the image of the spatial domain into Brightness Domain) using the following formula:
Capture.JPG
Does anyone know how to do this in Matlab??
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Rik
Rik 2019-3-12
So you want a window average, but with a maximum, with a 15-by-15-by-size(I,3) window?

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Image Analyst
Image Analyst 2019-3-12
编辑：Image Analyst 2019-3-12
The inner operation is dilation. The outer one is max. So simply dilate each color channel (take the max in a 15x15 window) with imdilate(). Like for the red channel.
dilatedRed = imdilate(redChannel, ones(3));
Same for blue and green, then recombine into an rgb image and use max
brightnessImage = max(rgbImage, [], 3);
A full demo is attached.
0000 Screenshot.png
But I don't recall seeing this formula you gave. Taking the max is what you do to get L in the RGB-to-HSL, or to get the V in RGB-to-HSV formula. But they don't do dilation first. Not sure what that gets you, do you know? Why not just do it without dilation and use the built-in rgb2hsv() function
hsvImage = rgb2hsv(rgbImage); % Convert to HSV color space.
brightnessImage = hsvImage(:, :, 3); % Extract the V channel.
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Image Analyst
Image Analyst 2019-3-13
Yes, except the order is switched. It is:
Imin = imerode(min(Ig, [], 3), ones(windowSize));
Where are you getting these formulas from? I can understand that the dark current signal might be estimated as the min of the red, green, and blue pixels in the demosaicing window but I really doubt any camera uses a demosaicing algorithm where the interpolated pixels are the uniformly weighted average inside a 15x15 window.
ghada sandoub
ghada sandoub 2019-3-13
I'm working now in researching and I get these formulas from a paper that interested in low light image enhancement. this paper uses the bright channel formula to estimate the insufficient illumination in the low light image.

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Rik
Rik 2019-3-12
Below you find 2.5 strategies. The first should be a lot faster, especially for larger images. With the last it is easier to see that the method returns the expected outcome.
clc
A=randi(20000,300,300,3);%random image
window=15;
%strategy 1:
tic
GetSingleLayerMax=@(A,window) nlfilter(A,[window window],@(x) max(x(:)));
maxIM=GetSingleLayerMax(A(:,:,1),window);
for n=2:size(A,3)
tmp=cat(3,maxIM,GetSingleLayerMax(A(:,:,n),window));
maxIM = max(tmp,[],3);
end
toc
%strategy 1b:
tic
GetSingleLayerMax2=@(A,window) colfilt(A,[window window],'sliding',@max);
maxIM_1b=GetSingleLayerMax2(A(:,:,1),window);
for n=2:size(A,3)
tmp=cat(3,maxIM_1b,GetSingleLayerMax2(A(:,:,n),window));
maxIM_1b = max(tmp,[],3);
end
toc
%strategy 2:
tic
maxIM2=zeros(size(A,1),size(A,2));
tmp=(1:window);tmp=tmp-mean(tmp);
tmp=ceil(tmp);%in case of even window size
[w_x,w_y,w_z]=meshgrid(tmp,tmp,1:size(A,3));
w=[w_x(:) w_y(:) w_z(:)];
for r=1:size(A,1)
for c=1:size(A,2)
%get all indices
tmp=w;
tmp(:,1)=w(:,1)+r;
tmp(:,2)=w(:,2)+c;
%remove invalid indices
tmp(any(tmp<1,2),:)=[];
tmp(tmp(:,1)>size(A,1),:)=[];
tmp(tmp(:,2)>size(A,2),:)=[];
%convert to linear indices
ind=sub2ind(size(A),tmp(:,1),tmp(:,2),tmp(:,3));
%compute the max
maxIM2(r,c)=max(A(ind));
end
end
toc
if isequal(maxIM,maxIM2) && isequal(maxIM_1b,maxIM2)
disp('everything finished succesfully')
else
disp('something went wrong (check for float rounding errors)')
end

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