Logical operations on matrix
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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?
1 个评论
Stephen23
2016-2-4
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.
采纳的回答
Kelly Kearney
2016-2-4
编辑：Kelly Kearney
2016-2-4
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);
3 个评论
Guillaume
2016-2-4
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
2016-2-4
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
2016-2-4
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.
更多回答（1 个）
Star Strider
2016-2-4
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;
1 个评论
Image Analyst
2016-2-4
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."
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