Best method for identifying the color

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i Venky
i Venky 2012-1-10
评论: DGM 2023-2-14
Hello. I have read a book in image processing. I have worked it out with matlab too. I want to find out the location (co-ordinates) of a particular color in the image. How should I do this? Which is the best method for color identification?
Thanks in advance.
  6 个评论
Walter Roberson
Walter Roberson 2019-11-20
移动:DGM 2023-2-14
You start by defining exactly what yellow means to you. For example is sunlight yellow, or is it yellow-green (as science tells us), or is it white?
Image Analyst
Image Analyst 2019-11-21
移动:DGM 2023-2-14
Try the Color Thresholder app on the Apps tab of the tool ribbon.
Or check out some demos in my File Exchange.

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回答(3 个)

Image Analyst
Image Analyst 2012-1-13
Color image analysis is not so easy, but it's one of my specialties as you can see from my logo and my File Exchange:
Check out the color segmentation demos I have there. There are three different ways of detecting colors. You can also specify how close you want the colors to be or how different you'll allow them to be from each other (like how much spread there is in the detected color gamut).

Jonathan Sullivan
Jonathan Sullivan 2012-1-11
编辑:Walter Roberson 2017-10-11
Think of the color of a pixel being a vector in a 3 dimensional space. What you want to do is find the angle of the color vector of the pixel to the color vector of the ideal by using the dot product.
For example:
blue = [0 0 255];
pixel = [10 10 200];
ang_thres = 25; % degrees. You should change this to suit your needs
ang = acosd(dot(blue/norm(blue),pixel/norm(pixel)));
isBlue = ang <= ang_thres; % Apply angle threshold
You might also want to apply a magnitude threshold (i.e. is the pixel dark enough). This would filter out any really faint colors (i.e. [0 0 1]);
For example:
blue = [0 0 255];
pixel = [10 10 200];
ang_thres = 25; % degrees. You should change this to suit your needs
ang = acosd(dot(blue/norm(blue),pixel/norm(pixel)));
mag_thres = 64; % You should change this to suit your needs
mag = norm(pixel);
isBlue = ang <= ang_thres & mag >= mag_thres; % Apply both thresholds
Hope this helps!
  2 个评论
Alejandro Hernandez6
thanks to all for this post. Anyway what means 'norm(blue)','norm(pixel)'? normalized ? My task is (given an image representing a floor plan) remove from it all green, red and blue pixels, considered noise, and later I will have to recognize all black segments to produce an floor plan.
Image Analyst
Image Analyst 2017-10-11
You're best off just attaching your image and telling us what you need to do, extract, or measure in a brand new question. Like what, exactly, is considered "noise."
You can also search the Image Analysis literature for papers on floor plans by clicking here.

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Walter Roberson
Walter Roberson 2012-1-11
Unfortunately, the hue-angle solution from Jonathan, and the hsv solution from Chandra, both will likely classify pink as being red, since pink is a saturated form of red. This is why it is important to define exactly what "red" and "yellow" and "blue" mean to you.
  3 个评论
Image Analyst
Image Analyst 2012-1-13
移动:DGM 2023-2-14
I ran it and it looks like it gives a red Chirp or sawtooth function. Not sure I understand.
DGM
DGM 2023-2-14
The example selects colors in a conical region around red.
% Set user parameters
mag_thres = 0.5; % Set magnitude threshold
ang_thres = 30; % Set angular threshold
r = cat(4,1,0,0); % Define "red"
% Create all colors
vv = single(linspace(0,1,200));
[R G B] = meshgrid(vv);
RGB = cat(4,R,G,B);
% Threshold
mag = sqrt(sum(RGB.^2,4));
ang = acosd(sum(bsxfun(@times,r,bsxfun(@rdivide,RGB,mag)),4));
f = ang < ang_thres & mag > mag_thres;
% Display Results
% this is a Nx1x3 stripe of colors within the selected volume
image(cat(3,R(f),G(f),B(f)))
% plot the slected volume
isosurface(R,G,B,f,0)
grid on
axis equal
ll = [0 1];
xlim(ll)
ylim(ll)
zlim(ll)
xlabel('R');
ylabel('G')
zlabel('B')
view(-73,35)
... but it doesn't seem generalized. The thresholds aren't relative to the specified color, so choosing any color other than red doesn't work.

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