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

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     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
     2     0     0     0     0     0     0     5

Answer 1

[Walter Roberson]

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)."

[Oman Wisni]

I already trying again, using NumLevels but the result same too. the result like this. please explain sir

[Walter Roberson]

>> graycomatrix(gray, 'numlevels', 3, 'graylimits', [])
ans =
     2     2     0
     1     2     1
     0     0     0

[Oman Wisni]

can you send the matrix of this answer ? and what function for graylimits? can give me more explain ? thanks sir

[Walter Roberson]

 m = graycomatrix(gray, 'numlevels', 3, 'graylimits', []);

if that is what you mean.

The graylimits option is described at https://www.mathworks.com/help/images/ref/graycomatrix.html#bu3j4f2-1-GrayLimits

[Oman Wisni]

Yes, thanks sir I will trying. Sir, if I want to find graycomatrix from image leaf how to declarad? I dont know how many NumLevel in image gray leaf

[Walter Roberson]

Finding out how many levels you need for accurate enough classification for your purposes, is part of the research process.

[Oman Wisni]

Yes sir. Thanks. But I though use bit of the gray image. For example I have image.bmp, the bit of my image is 8 bit so I though the range value of NumLevel 256 ?

[Walter Roberson]

For 8 bit grayscale, using more than 256 will not help. For the purpose of matching textures, it is likely that you should be using a lower count, to take into account noise and lighting differences.

[Oman Wisni]

Here I have some code using formula to find GLCM but the result was 0.

in below I attach an image and my code :

S = size(gray, 4);

if any(gray(:))
    prob = gray./sum(gray(:));
end

prob_norm = gray ./ sum(gray(:));

u = mean2(prob_norm);
py = sum(prob_norm,1);
px = sum(prob_norm,2);

contr = 0;
energ = 0;
homom = 0; 
corr = 0;

for i = 1:S
    for j = 1:S

          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    = i*j*prob(i,j);
       end
  end

%Correlation
ux = mean(px); sx=std(px);
uy = mean(py); sy=std(py);
corr =(sum(co(:))-(ux*uy))/(sx*sy);

      Stat.Contrast = contr;
      Stat.Homogeneity = homom;
      Stat.Energy = energ;
      Stat.Correlation = corr;

[Walter Roberson]

prob = gray./sum(gray(:));

if gray is integer data type, the result of gray ./ a double precision number would be the integer data type, which is probably mostly going to be 0 but might have some 1 as well.

Try

prob = double(gray) ./ sum(gray(:));

[Oman Wisni]

Only that I change? How about the size and prob_norm?

The result still 0

[Walter Roberson]

I do not know. I am not going to look at this now; I am resting after some surgery.

[Oman Wisni]

Oohh Im so sorry sir. I dont know. Yes sir, its okay I will asking another one.. Thanks for the answer and get well soon.

[Walter Roberson]

Your prob and prob_norm are the same. You should be using double(gray) for both.

Your loops only examine the image at (1:5, 1:5), which is an area that is only uniform background.

Your image is a pseudocolor image. To read it properly you should be using

[img_as_ind, cmap] = imread('6_grayscale.bmp');
img_as_rgb = ind2rgb(img_as_ind, cmap);
gray = rgb2gray(img_as_rgb);

which would happen to leave the values as double and so the double(gray) would not be needed.

[Oman Wisni]

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

[Walter Roberson]

I am not familiar with those formulas.

[Oman Wisni]

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]

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

Answer 2

[sam alsalihy]

as the following

1 2 7 6 10

7 6 2 7 4

1 2 4 6 2

2 2 2 4 1

then the glcm matrix calculated from SI

1 2 3 4 5 6 7 8 9 10

1: 0 2 0 0 0 0 0 0 0 0

2: 0 2 0 2 0 0 2 0 0 0

3: 0 0 0 0 0 0 0 0 0 0

4: 1 0 0 0 0 1 0 0 0 0

5: 0 0 0 0 0 0 0 0 0 0

6: 0 2 0 0 0 0 0 0 0 1

7: 0 0 0 1 0 2 0 0 0 0

8: 0 0 0 0 0 0 0 0 0 0

9: 0 0 0 0 0 0 0 0 0 0

10: 0 0 0 0 0 0 0 0 0 0