The right way to find the number of nonzero elements (in general) is to use the nnz() function; using sum() also works in this particular case but will fail if there are numbers other than zero and one in the matrix used. Therefore to calculate the total element count, nonzero element count, and ratio, use code like this:

Copyx = [1 1 1 1 1 1 0 0 1 0];
nonzeroes = nnz(x);
total = numel(x);
ratio = nonzeroes / total;

correct me if I'm wrong but it sounds like you want the number of occurrences of the numbers in your vector, here's an alternative if that is the case:

arr=[1 2 2;3 3 3;5 0 0;0 0 0]; % example array where 1,2,3 occur 1x,2x,3x and 5=>1x, 0=>5x
[x(:,2),x(:,1)]=hist(arr(:),unique(arr(:))); 

outputs sorted category as first column, occurrences as 2nd column:

x =

     0     5
     1     1
     2     2
     3     3
     5     1

As you seem to be interested in checking it manually, perhaps to review matlab code, here is the trick:

• 1:8 has 8-1+1 = 8 elements
• -5:5 has 5--5+1 = 11 elements
• -10:5 has 5--10+1 = 16 elements

So the result is:

The matrix has 8*11*16 = 1408 elements in total.

Here's a list of all the ways I could think of to counting unique elements:

M = randi([1 7], [1500 1]);

Option 1: tabulate

t = tabulate(M);
counts1 = t(t(:,2)~=0, 2);

Option 2: hist/histc

counts2_1 = hist( M, numel(unique(M)) );
counts2_2 = histc( M, unique(M) );

Option 3: accumarray

counts3 = accumarray(M, ones(size(M)), [], @sum);
%# or simply: accumarray(M, 1);

Option 4: sort/diff

[MM idx] = unique( sort(M) );
counts4 = diff([0;idx]);

Option 5: arrayfun

counts5 = arrayfun( @(x)sum(M==x), unique(M) );

Option 6: bsxfun

counts6 = sum( bsxfun(@eq, M, unique(M)') )';

Option 7: sparse

counts7 = full(sparse(M,1,1));