Counting the neighbors in matrix

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Aravin
Aravin 2012-8-7
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 :-(
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Aravin
Aravin 2012-8-13
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
Matt Fig 2012-8-13
编辑：Matt Fig 2012-8-13
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.

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采纳的回答

Matt Fig
Matt Fig 2012-8-8
编辑：Matt Fig 2012-8-8
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];

更多回答（3 个）

Sean de Wolski
Sean de Wolski 2012-8-7
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
Andrei Bobrov
Andrei Bobrov 2012-8-7
编辑：Andrei Bobrov 2012-8-8
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
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Aravin
Aravin 2012-8-11
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
Sean de Wolski 2012-8-13
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 :)

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Teja Muppirala
Teja Muppirala 2012-8-8
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);

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