You extract
Klocal=K(CentroidTermY(ind)-2:CentroidTermY(ind)+2,CentroidTermX(ind)-2:CentroidTermX(ind)+2);
Klocal(2:end-1,2:end-1)=0;
so Klocal is extracted as an array with 5 rows and 5 columns, and you then zero out the lower right hand side. That leaves the top row and left column as being potentially non-zero -- Klocal(1,:) and Klocal(:,1) might be non-zero, a total of 9 different locations that are potentially non-zero.
That asks for the location of all of the non-zero values in the 5 x 5 array that resulted above. With 9 different locations that were not explicitly 0, you could get anywhere between 0 and 9 results. The i and j are going to be returned as column vectors in this case.
OrientationTerm(ind,1)=Table(i,j);
With ind being scalar and 1 being scalar, the left hand side names exactly one output location. The right hand side, on the other hand, names either 0, 1, 4, 9, 16, 25, 36, 49, 64, or 81 locations in Table, because when you index with two vectors of values, the output corresponds to taking all combinations of indices.
If you wanted to pull corresponding individual entries out of Table instead of rows and columns, you would use
temp = Table( sub2ind(size(Table), i, j) );
sub2ind would take those column vectors of indices and create individual indices from each corresponding row, so the above would end up designating 0 through 9 output locations.
You would then have to figure out how you wanted to store the 0 through 9 entries into OrientationTerm
Perhaps what you want is
[i,j] = find(Klocal, 1);
if ~isempty(i)
OrientationTerm(ind,1) = Table(i,j);
else
OrientationTerm(ind,1) = nan;
end
This finds one of the non-zero entries, tending with it tending to prefer to take from the first column (that you did not zero out) before it would take from the rest of the first row, because of the order that arrays are scanned.
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