Logical operations on matrix

Hello,

I thought i had this figured but i am having trouble with logical operations on a matrix. The code is:

% Quantization stepsize 
hQ = 12;

% Quantization levels 
hQInt = [0:hQ:360];

% Quantization matrix 
hueQuant = zeros(600,800);

for i=5:5
    hueQuant(hQInt(i)< hue >hQInt(i+1)) = 1;
end

The idea: I am making quantization intervals with hQInt. I want to use a logical operation in the "for-loop", and i thought this statement said - "The index in hue where hue is larger than hQInt(i) and less than hQInt(i+1) is set to 1 in the matrix hueQuant". Just to mention it, the iteration index is set to 5 for test purpose.

The question: Am i wrong or is it possible this way? Any suggestions to how it can be solved without a massive for loop code?

[Stephen23]

Vegard B's "Answer" moved here:

Thanks to both of you for the reply! The solution by Kelly Kearney removed the need to loop and was chosen.

Answer 1

[Kelly Kearney]

Assuming I'm understanding your intent, a better way to do this would be with histc. No loop necessary:

hQ = 12;
hQInt = [0:hQ:360];

hue = rand(600,800)*360;

[~,hueQuant] = histc(hue, hQInt);

[Guillaume]

discretize introduced in R2015a is even more suited for this.

hueQuant = discretize(hue, [hQInt, inf]); %discretize treats the last edge differently than histc

[Kelly Kearney]

Oh, nice! I hadn't noticed that addition, and both the last-edge treatment and the ability to specify which bin side is the included in the bins are much-needed features for me.

[Image Analyst]

That's probably what is wanted. I never was able to figure out why they wanted to set hueQuant equal to 1 essentially everywhere. Both in code, and verbally they said that. That doesn't make sense to me.

Answer 2

[Star Strider]

Assuming I understand correctly what you want to do, this statement:

hueQuant(hQInt(i)< hue >hQInt(i+1)) = 1;

would likely do what you want if you restated it as:

hueQuant((hQInt(i)< hue) & (hue >hQInt(i+1))) = 1;

[Image Analyst]

Of course hue would have to be defined first. And it's funny how they talk about a massive for loop when none of the arrays or sizes there, like 5 or 600*800, constitute "massive."