MATLAB: Finding the address of an element in a multi-dimensional array

ind2sublinear addressmatrixmultidimensional arraysubscripts

function [cMatrix] = makeC (n,f)
%n is the degree of the polynomial, f is the number of factors (dimensionality)
A = [];
if f == 1
    cMatrix = 2*rand(n+1,1)-1;
else
    for i = 1:f
        A(1,i) = n+1;
    end
    cMatrix = rand(A);
end
for i = 1:numel(cMatrix)
    coords = ind2sub(size(cMatrix),i);
    coordSum = 0;
    for ii = 1:numel(coords)
        check = check + coords(ii);
        if coordSum < f(n+1) - n
            cMatrix(i) = 0;
        end
    end
end
The problem is that ind2sub (a built-in matlab function that converts a linear index into subscripts) does not produce the correct number of coordinates on its own. I have to say something like:
[x , y] = ind2sub([2,2],3)
to make [x,y] become [1 , 2]
but the number of dimensions in the multidimensional array is at the discretion of the user. So I need the output of ind2sub to be a 1xf matrix where f is the number of dimensions named coords. Currently coords is just the linear address! I am new to matlab (like as of yesterday) so please be thorough.
Possibly irrelevant information regarding the purpose of the function: generates a matrix of random coefficients for an nth order polynomial. The problem is that if you want a 2nd order polynomial with 2 factors (3×3 matrix) you don't want terms like x^2*y^2 because that is a quartic term. So the coef matrix for 2nd order 2 factors would look like:
       x^2    x     c
y^2     0     0   rand#
 y      0   rand# rand#
 c    rand# rand# rand#
The first half of the code makes the appropriately sized matrix the second half iterates through and finds where the zeros go.

Best Answer

It is easy to create a function that will return the subscript indices in a vector:
function vec = ind2vec(siz,ndx)
tmp = [1,cumprod(double(siz(1:end-1)))];
for k = numel(siz):-1:1,
    nxt = 1 + rem(ndx-1, tmp(k));
    vec(k) = 1 + (ndx - nxt)/tmp(k);
    ndx = nxt;
end
end
And demonstrated:
>> ind2vec([2,2],3)
ans =
     1     2
>> ind2vec([2,2,3,4],3)
ans =
     1     2     1     1
>> ind2vec([2,2,3,4],32)
ans =
     2     2     2     3
Related Question