I have an array A which has a size [N1 N2 N3 3 3]. Array A can be considered as a rectangular box in three dimensions that for each of its element, an array of size [3 3] is defined, and here I call it B. In other words, B with size [3 3] is defined for each point in the domain.
I am trying to subdivide the rectangular box into small cubes of size d×d×d, where d is an even number. Also, there should be 50% overlap between adjacent cubes and . For example, let's say:
A; % size(A) = [14 16 20 3 3];
d = 4;
With the above example, the sampling will be as follows. The cubes along with with first dimension of A, i.e., 1:14 will include the following elements of A:
cube1: 1, 2, 3, 4
cube2: 3, 4, 5, 6
cube3: 5, 6, 7, 8
cube4: 7, 8, 9, 10
cube5: 9, 10,11,12
cube6: 11,12,13,14
along the second dimension of A, i.e., 1:16, we have:
cube1: 1, 2, 3, 4
cube2: 3, 4, 5, 6
cube3: 5, 6, 7, 8
cube4: 7, 8, 9, 10
cube5: 9, 10,11,12
cube6: 11,12,13,14
cube7: 13,14,15,16
And finally along the third dimension of A, i.e., 1:20:
cube1: 1, 2, 3, 4
cube2: 3, 4, 5, 6
cube3: 5, 6, 7, 8
cube4: 7, 8, 9, 10
cube5: 9, 10,11,12
cube6: 11,12,13,14
cube7: 13,14,15,16
cube8: 15,16,17,18
cube9: 17,18,19,20
Now Anew will have a size [6 7 9 3 3] and the value at each of its element will be the summation of each element of B over all the element of A that are in each cube. For example, Anew(1, 4, 9, 1, 1) is actually:
A(elements of cube1, elements of cube4, elements of cube9, 1, 1)
or in other words, A(1:4, 7:10, 17:20). Therefore:
Anew(1, 4, 9, 1, 1) = 0;
for i = 1:4 % cube1 along with first dimension uses 1, 2, 3, 4
for j = 7:10 % cube4 along with the second dimension uses 7, 8, 9, 10
for k = 17:20 % cube9 along with the third dimension uses 17, 18, 19, 20
Anew(1, 4, 9, 1, 1) = Anew(1, 4, 9, 1, 1) + A(i, j, k, 1, 1);
end
end
end
In the above for loop, A(i, j, k, 1, 1) is B(1, 1) at point (i, j, k) in the original domain. Could someone help me how I should approach in a vectorize fashion?
The rows of cube1,2,etccan be given by 2*n-1:2*n+2 where n is the row number. this will be stored as an anonymous function to simplify the later code...
ind = @(n) 2*n-1 : 2*n+2
Vectorising the loop in the question, for point(ii,jj,kk) the value of B(ll,mm) a possible solution is as follows:
sum(reshape(A(ind(ii),ind(jj),ind(kk),ll,mm),1,[]))
where the above code gets all required elements, reshapes into a column and then sums
To generate values for all ii,jj,kk,ll,mm combinations
sA=size(A);
nii=sA(1)/2-1; njj=sA(2)/2-1; nkk=sA(3)/2-1; nll=sA(4); nmm=sA(5) %Sizes for grid
[ii,jj,kk,ll,mm]=ndgrid(1:nii,1:njj,1:nkk,1:nll,1:nmm); %Creates grid
Anew=arrayfun(@(ii,jj,kk,ll,mm)sum(reshape(A(ind(ii),ind(jj),ind(kk),ll,mm),...
1,[])),ii,jj,kk,ll,mm); %Apply function to grid