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Counting the neighbors in matrix

Asked by Aravin on 7 Aug 2012

Hi everyone,

I have matrix R in which each element is integer of range [0 7]. I want to calculate the count of neighbor, let say for pixel value 0 how many times it has the neighbor 1, 2, ..., 7. every pixel has eight neighbors, for example

[n1 n2 n3
 n4 x  n5
 n6 n7 n8]

where pixel x have eight neighbors. As a resultant I think we will have an other matrix of 8 x 8, or we can say 8 histograms as we have eight possible values in given matrix R.

I hope I have explained what I want :-(

9 Comments

Image Analyst on 12 Aug 2012

Granted, their graycomatrix is very confusing and has some weird defaults, and you have to do things to get it to do what you'd intuitively want it to do. For example, it only looks at pairings to the right of the pixel, not all around at all 8 neighbors (unless you pass in the correct "offset" parameter). Matt, try this code and see if you get the tables they show in their explanatory diagram:

m = [1 1 5 6 8; 2 3 5 7 1; 4 5 7 1 2; 8 5 1 2 5]
glcm = graycomatrix(m, 'GrayLimits', [min(m(:)) max(m(:))])
Aravin on 13 Aug 2012

Hi Matt,

Here is the one line solution I did for my problem.

final_H= graycomatrix(x, 'offset', [0 1; -1 1; -1 0; -1 -1; 0 -1; 1 -1;1 0;1 1],'NumLevels',15);

I will have 8 number of channels, which I have sum for final result.

Matt Fig on 13 Aug 2012

Thanks, IA. Indeed you are correct. The illustration I was looking at in the doc seemed to imply it was showing graycomatrix(I).

I see now the difference between what this function gives and what I (and others) showed below. If we have:

x = randi([0 7],1000,1000);

Then to get the same result as:

G = sum(graycomatrix(x,...
        'graylimits',[0 7],...
        'Offset',[0 1; 0 -1;-1 1; -1 0; -1 -1;1 -1;1 0;1 1]),3)

I would modify my code to:

R = ones(3);
R(5) = 0;
for ii = 7:-1:0
    I = conv2(single(x==ii),R,'same');
    for jj = 0:7
        M(ii+1,jj+1) = sum(I(x(:)==jj));
    end
end

Thus M and G are equal. M took half the time to compute, but the code is definitely longer! Note that the calculation of M could perhaps be made more efficient by storing the x(:)==jj in a cell outside the loop.

Anyway, I'm glad Aravin got what was required.

Aravin

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4 Answers

Answer by Matt Fig on 8 Aug 2012
Edited by Matt Fig on 8 Aug 2012
Accepted answer

If you don't have the image processing toolbox, or if you want your code to be useful to those who don't, this is just as fast. Note that the example only looks for the neighbors of the number 7. You can put this into a loop as needed. I assume your matrix is named x.

H = single(x==7);
H = logical(conv2(H,ones(3),'same'))~=H;
H = x(H(:));
Y = unique(H);   
H = histc(H,Y); 
H = [Y H]; 

1 Comment

Aravin on 11 Aug 2012

Thanks Matt...

Matt Fig
Answer by Sean de Wolski on 7 Aug 2012

I would take an image processing approach:

x = zeros(5); %sample matrix
x(3:5,3:5) = [1 2 1;1 7 2;3 4 1];
x(5) = 7;
BW = x==7; %We'll count neighbors of 7s
M  = xor(imdilate(BW,ones(3)),BW); %neighbors of 7s
uv = unique(x(M)); %unique neighbors
n = histc(x(M),uv); %count unique neighbors
[uv n] %display number of unique values to occurences

2 Comments

Matt Fig on 8 Aug 2012

Nice!

Aravin on 11 Aug 2012

Thanks Sean... Really nice solution.

Sean de Wolski
Answer by Andrei Bobrov on 7 Aug 2012
Edited by Andrei Bobrov on 8 Aug 2012
A = randi([0 7],10);
A1 = nan(size(A)+2);
A1(2:end-1,2:end-1) = A;
k = (0:7)';
n = numel(k);
out = zeros(n);
d = true(3);
d(5) = false;
for i1 = 1:n
    [ii,jj] = find(A1 == k(i1));
    p = zeros(n,1);
    for i2 = 1:numel(ii)
        q = A1(ii(i2) + (-1:1),jj(i2) + (-1:1));
        p = p + histc(q(d&(~isnan(q))),k);
    end
    out(:,i1) = p;
end

variant based on the idea of ​​Sean and Matt:

k = (0:7)';
out = zeros(numel(k));
for a = 1:numel(k)
    out(:,a) = histc(x(bwdist(x == k(a),'chessboard') == 1),k);
end

3 Comments

Aravin on 11 Aug 2012

Thanks Andrei Bobrov :-)

Aravin on 11 Aug 2012

Indeed, this is the perfect solution to my problem. Matts and sean Solution doesn't include the count of searching key value. In their given examples, if we are seaching 7, then their solution doesn't count the 7 as neighbor, but your solution does.

Thanks Andrei.

Sean de Wolski on 13 Aug 2012

Our solutions could count the 7 as a nieghbor! Just remove the xor() from mine or the ~= from Matt's. We were intentionally not counting the 7, but if you want to count it it makes our solutions even simpler :)

Andrei Bobrov
Answer by Teja Muppirala on 8 Aug 2012
R = randi([0 7],100); % Input
H = zeros(8); % The 8x8 Matrix
for k = 0:7
    C = conv2(double(R == k),[1 1 1; 1 0 1; 1 1 1],'same');
    H(:,k+1) = accumarray(1+R(C~=0),nonzeros(C),[8 1]);
end
bar(H);

0 Comments

Teja Muppirala

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