Compressed Ratio in HSL color space

Question

Nour George 2020-7-13

0
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https://ww2.mathworks.cn/matlabcentral/answers/564062-compressed-ratio-in-hsl-color-space

编辑： Nour George 2020-7-13

I have This three Question :

1)Read the Image ('flag.png')

2) convert it into ondex type and display it

3)Compression Ration ((Assignement))

Assignement:

HSI color space

*I solved the first and the second Question But for the Third Question I convert RGB to HSL to Compressed Ratio, The Question is How can i compressed Ration after converting to HSL ? AND is the converting to HSL correct ?

figure

%1

image_mat = imread('picture1.png');

imshow(image_mat)

%-----------------------End the first Question

%2

[image_mat, cmap] = imread('picture1.png');

if isempty(cmap)

if ndims(image_mat) > 2

image_rgb = image_mat;

else

image_rgb = repmat(image_mat, [1 1 3]);

end

else

image_rgb = ind2rgb(image_mat, cmap);

end

imshow(image_rgb)

%-----------------------End the second Question

%3

function hsl=rgb2hsl(image_rgb)

%Converts Red-Green-Blue Color value to Hue-Saturation-Luminance Color value

%

%Usage

% HSL = rgb2hsl(RGB)

%

% converts RGB, a M X 3 color matrix with values between 0 and 1

% into HSL, a M X 3 color matrix with values between 0 and 1

if nargin<1,

error('Too few arguements for rgb2hsl');

return;

elseif nargin>1,

error('Too many arguements for rgb2hsl');

return;

end;

if max(max(image_rgb))>1 | min(min(image_rgb))<0,

error('RGB values have to be between 0 and 1');

return;

end;

for i=1:size(image_rgb,1),

mx=max(image_rgb(i,:));%max of the 3 colors

mn=min(image_rgb(i,:));%min of the 3 colors

imx=find(image_rgb(i,:)==mx);%which color has the max

hsl(i,3)=(mx+mn)/2;%luminance is half of max value + min value

if(mx-mn)==0,%if all three colors have same value,

hsl(i,2)=0;%then s=0 and

hsl(i,1)=0;%h is undefined but for practical reasons 0

return;

end;

if hsl(i,3)<0.5,

hsl(i,2)=(mx-mn)/(mx+mn);

else

hsl(i,2)=(mx-mn)/(2-(mx+mn));

end;

switch(imx(1))%if two colors have same value and be the maximum, use the first color

case 1 %Red is the max color

hsl(i,1)=((image_rgb(i,2)-rgb(i,3))/(mx-mn))/6;

case 2 %Green is the max color

hsl(i,1)=(2+(image_rgb(i,3)-rgb(i,1))/(mx-mn))/6;

case 3 %Blue is the max color

hsl(i,1)=(4+(rgb(i,1)-rgb(i,2))/(mx-mn))/6;

end;

if hsl(i,1)<0,hsl(i,1)=hsl(i,1)+1;end;%if hue is negative, add 1 to get it within 0 and 1

end;

hsl=round(hsl*100000)/100000; %Sometimes the result is 1+eps instead of 1 or 0-eps instead of 0 ... so to get rid of this I am rounding to 5 decimal places)

end

%-----------------------End the third Question