Hi, I want ask you why the size of graycomatrix 8 x 8 ? that size default 8 x 8 or how ?

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Oman Wisni 2018-10-16

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回答： sam alsalihy 2020-5-16

采纳的回答： Walter Roberson

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sir I example matrix gray = [0 1 2 2; 1 1 2 1; 1 0 2 0;]; but after I m=graycomatrix(gray); the result like this :

m =

   0     0     0     0     0     0     2
   0     0     0     0     0     0     0
   0     0     0     0     0     0     0
   0     0     0     0     0     0     0
   0     0     0     0     0     0     0
   0     0     0     0     0     0     0
   0     0     0     0     0     0     0
   0     0     0     0     0     0     5

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Walter Roberson 2018-10-16

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It is because NumLevels defaults to 8

https://www.mathworks.com/help/images/ref/graycomatrix.html#bu3j4f2-1-NumLevels

"Number of gray levels, specified as an integer. For example, if NumLevels is 8, graycomatrix scales the values in I so they are integers between 1 and 8. The number of gray-levels determines the size of the gray-level co-occurrence matrix (glcm)."

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Oman Wisni 2018-10-20

编辑：Oman Wisni 2018-10-20

在 MATLAB Online 中打开

Thanks for the answer sir, have you recovered?

so I should use that code before counting ? I thought only use graycomatrix. Before I already trying sir. The result from graycoprops() and manuall counting already same, but only correlation different. I will attached the code so you can understand I mean, cause I can't properly explain. May you can trying to run this code, using graycoprops(); and formula I give.

glcm = [2 1 2; 1 1 1; 1 2 3; 2 2 2;]; 
m = graycomatrix(glcm, 'numlevels',3, 'GrayLimits', []);
gl = graycoprops(m);
prob_norm = m ./ sum(m(:));
%u = mean2(prob_norm);
%py = sum(prob_norm);
%px = sum(prob_norm);
contr = 0;
energ = 0;
homom = 0; 
co    = 0;
%corr = 0;
for i = 1 : 3
    for j = 1 : 3 
          energ = energ + prob_norm(i,j)^2;
          contr = contr + prob_norm(i,j).*(i-j)^2;
          homom = homom + prob_norm(i,j)/(1+(i-j)^2);
          co    = co + prob_norm(i,j);
      end
  end
ui = 0;
uj = 0;
si = 0;
sj = 0;
for i = 1
    for j = 1
        ui = ui + i.* prob_norm(i,j);
        a = i - ui;
        uj = uj + j.* prob_norm(i,j);
        b = j - uj;
        stdi = si + prob_norm(i,j).*(i-ui);
        stdj = sj + prob_norm(i,j).*(j-uj);
    end
end
corr = co.*(a.*b)/(stdi.*stdj);
%Correlation
%ux = mean(px); sx=std(px);
%uy = mean(py); sy=std(py);
%corr =(sum(co(:))-(ux*uy))/(sx*sy);
%Correlation
%ux = mean(px); sx=std(px);
%uy = mean(py); sy=std(py);
      Stat.Contrast = contr;
      Stat.Homogeneity = homom;
      Stat.Energy = energ;
      Stat.Correlation = corr;
disp ('========== Fitur Tekstur GLCM ==========')
disp(strcat(['Contrast = ',num2str(contr)]))
disp(strcat(['Homogenitas = ',num2str(homom)]))
disp(strcat(['Energy = ',num2str(energ)]))
disp(strcat(['Correlation = ',num2str(corr)]))

For image leaf which I attached before I do same like matrix above, only I change in this code

glcm = [gray]; % mean the image 6_grayscale.bmp is gray.
m = graycomatrix(glcm);
gl = graycoprops(m);
For i = 1:8;
.....
For j = 1:8;

My question now, Is this formula correct for me to apply in the picture of my leaves? indeed the results are the same, but I doubt using graycomatrix (); and in the correlation formula, I'm still wrong sir, the results are not appropriate, please help.

if only using graycomatrix () default for my leaves is enough without having to convert to double(gray), is that value enough for me to use feature extraction? Thank you sir

Oman Wisni 2018-10-22

编辑：Oman Wisni 2018-10-22

Its okay sir, that formula I got from journal and the theory from haralick. Thanks for answers sir I already get the result and same with graycoprops () :)

sam alsalihy 2020-5-16

it is by defalt the output scale to 8. It ismean that the value of input image will be converted from its range to the range between 0-8 then calculate the glcm matrix

for example if you set image pp=[1 2 5 4 7;5 4 2 5 3; 1 2 3 4 2;2 2 2 3 1 ]

and [GLCMS,SI] = graycomatrix(pp,'NumLevels',7,'G',[]) so the NumLevels=7 it s in the same range ofthe pp image thus will convert matrix SI

1 2 5 4 7

5 4 2 5 3

1 2 3 4 2

2 2 2 3 1

it mean same orginal image. but for example the NumLevels=10 it will convert to range 1-10, thus the SI matrix will be

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