How to calculate the entropy of a portion of image as in the following formula?

Hello everyone, I am new here on matlab. I should calculate the subset entropy of portion of image, defined as in the formula (4). Could someone please tell me how to write a routine that calculates the subset entropy using matlab please.

Thank you very much in advance.

[Walter Roberson]

nlfilter

[Simone]

@Walter Roberson How can I use "nifilter" to implement the formula (4)?

Answer 1

[Bruno Luong]

Please try this

A=imread('ngc6543a.jpg');;
A=double(A(1:500,:,:));
A=sum(A,3);

M = 11; N = 11;

% Build 8-neighbor
[di,dj] = ndgrid(-1:1);
di = di(:);
dj = dj(:);
di(5,:) = []; dj(5,:) = []; % remove (0,0) shift
A8 = zeros([size(A),8]);
for i=1:8
    A8(:,:,i) = A(min(max(di(i)+(1:end),1),end), ...
        min(max(dj(i)+(1:end),1),end));
end
dA = sum(abs(A8-A),3);
depth = 24+zeros(size(A));
K = ones(M,N);
delta = conv2(dA,K,'same')./(2.^depth.*conv2(ones(size(dA)),K,'same'));

subplot(2,1,1);
imagesc(A)
colormap gray
subplot(2,1,2);
imagesc(delta);

[Simone]

hi @Bruno Luong, thanks for the reply. The entropy subset should be a scalar and not a matrix, like the one you defined. How can this be fixed?

[Bruno Luong]

A set (S) cannot a scalar, otherwise I don't understand what is the set all those years.

I my code the set S is MxN (11 x 11) pixels centered about all given pixel. If you want just one just extract one value in the image delta of my code.

[Simone]

but the subset entropy, defined as in formula (4) is given by a sum of values obtained from an array, so it is a scalar. Or maybe I don't understand what you mean

[Bruno Luong]

I compute as many delta as pixel of the original image (the set S is centered at the given pixel), if you want one delta corresponds to one S just pick one pixel of the image delta, that is a scalar. for example delta_100_200

A=imread('ngc6543a.jpg');;
A=double(A(1:500,:,:));
A=sum(A,3);
M = 11; N = 11;
% Build 8-neighbor
[di,dj] = ndgrid(-1:1);
di = di(:);
dj = dj(:);
di(5,:) = []; dj(5,:) = []; % remove (0,0) shift
A8 = zeros([size(A),8]);
for i=1:8
    A8(:,:,i) = A(min(max(di(i)+(1:end),1),end), ...
        min(max(dj(i)+(1:end),1),end));
end
dA = sum(abs(A8-A),3);
depth = 24+zeros(size(A));
K = ones(M,N);
delta = conv2(dA,K,'same')./(2.^depth.*conv2(ones(size(dA)),K,'same'));

% S = pixels A(95:105,195:205)
delta_100_200 = delta(100,200)

delta_100_200 = 6.6008e-08

[Simone]

Ok, I get what you mean. but i have another doubt. Other studies report that the entropy subset is of the order 0.3-0.5 but in this case the values of the delta matrix are in the order of 10^-8. How is it possible?

[Bruno Luong]

Ask people who claim it, here my value is smal becaise the depth (beta) is 24 as in your example of code.

1/2^24

ans = 5.9605e-08

You have all the values in my code, why can't you check it.

I simply implement what you wrote above as formula, I don't know this specific formula, the paper and people who wrote it.

What I know is entropy never have strict unit definition, as long as it's defined consistently across the usage for comparison.

[Simone]

You're right, I hadn't thought of that. A thousand thanks @Bruno Luong

Answer 2

[Image Analyst]

What about entropy

e = entropy(grayImage(row1:row2, col1:col2))

[Simone]

Hi @Image Analyst, the the way the entropy function is calculated is different from formula (4). I wrote this routine, do you think it will work?

A1=imread('IMG_111.png');
A=double(A1);
g=rgb2gray(A);
   
    x=2000 %coordinate of subset center
    y=2000 %coordinate of subset center
    j=60   %subset radius
    A1=A([(x-(j-1)):(x+j)],[(y-(j-1)):(y+j)]); 
    xx=((j*2)-9)
    for k=8:1:xx
    for l=8:1:xx
    A2=A1([(k-7):(k+9)],[(l-7):(l+9)]);
    for i=1:1:17
        for j=1:1:17 
          
           
            Ii=A2([i],[j]);
            row=size(A2,1);
            column=size(A2,2);
            Ip=A2([(row-1)/2],[(column-1)/2]);
            c=abs(Ip-Ii);
            C(i,j)=c;
            S = sum(C(:));
            
        end
    end
    D(k,l)=S;
    end 
    end
   S1 = sum(D(:)); 
   subsetentropy=S1/((2^24)*(j*2*j*2))