accessing all elements along the diagonals and rows (simple backprojection algorithm for image reconstruction)
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Hi, I'm wondering how to effectively sum all elements along every diagonal and row that touches an element in a given matrix. For example, in the matrix below
1 2 3
4 5 6
7 8 9
I would like to compute 1 + (1+2+3)/3 + (1+5+9)/3 + (1+4+7)/3. This value should replace the element 1 at (1,1). To replace the element 2 at (1,2), it would be (2+4)/2 + (2+6)/2 + (1+2+3)/3 + (2+5+8)/3. Similarly, element 6 at (2,3) should be (2+6)/2 + (6+8)/2 + (4+5+6)/3 + (3+6+9)/3.
It's a bit like a Sudoku calculation. The algorithm needs to be flexible for any square matrix, of any size (my data set is 360x360). Any help is much appreciated!
2 Commenti
KSSV
il 19 Ott 2018
You can access diagonal elements using diag. Row elements by A(i,:).
Image Analyst
il 19 Ott 2018
For a360x360 matrix, does the sum just go over the 3 nearest elements inside the matrix, or does it go all the way to the outer edge of the matrix?
Explain why you want to do this thing. What's the use case?
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Più risposte (3)
Image Analyst
il 19 Ott 2018
0 voti
You can do this with conv2(). If you can't figure it out, let me know.
1 Commento
Andrei Bobrov
il 19 Ott 2018
Modificato: Andrei Bobrov
il 19 Ott 2018
A = [1 4 7 10 13;
2 5 8 11 14;
3 6 9 12 15];
s = size(A);
n = ones(s(1),1);
a = n*sum(A)/s(1);
b = sum(A,2)*ones(1,s(2))/s(2);
k = sum(triu(rot90(triu(ones(s + [0, s(1)-1])),2)));
c = sum(spdiags(A))./k;
d = sum(spdiags(rot90(A)))./k;
dlr = full(spdiags(n*c,1-s(1):s(2)-1,s(1),s(2)));
drl = flip(full(spdiags(n*d,1-s(1):s(2)-1,s(1),s(2))),1);
out = a + b + dlr + drl;
variant by Guillaume's idea
s = size(A);
n = min(s)*2-1;
o = ones(s);
Iud = eye(n);
Idu = flip(Iud,2);
nn = conv2(o,Iud,'same');
out1 = conv2(A,Iud,'same')./nn + ...
conv2(A,Idu,'same')./flip(nn,2) + ...
o(:,1)*sum(A)/s(1) + sum(A,2)*o(1,:)/s(2);
Bruno Luong
il 19 Ott 2018
Modificato: Bruno Luong
il 19 Ott 2018
OK, here is the conv method as proposed by IA and started by Guillaume, it works for rectangular input array as well. Some convolution kernels are computed larger than necessary for clarity.
A = [1 2 3
4 5 6
7 8 9];
[m,n] = size(A);
p = 2*m-1;
q = 2*n-1;
r = (1:p)';
c = (1:q)';
d = min(m,n)-1;
k = (-d:d)';
Kh = accumarray([m+0*c, c],1,[p,q]);
Kv = accumarray([r, n+0*r],1,[p,q]);
Kd = accumarray([m+k, n+k],1,[p,q]);
Ka = accumarray([m+k, n-k],1,[p,q]);
I = ones(size(A));
cfun = @(K) conv2(A,K,'same')./conv2(I,K,'same');
% alternative way of calculation of h and v:
% [h,v] = ndgrid(mean(A,2),mean(A,1))
h = cfun(Kh);
v = cfun(Kv);
d = cfun(Kd);
a = cfun(Ka);
Result = h+v+d+a
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