Counting occurrences of a pointer

n=5;

h=zeros(n,1);
u=ceil(rand(n,1)*n);   % random sample on (1,n) with replacement
h(u) = h(u) +1;

u = [3 4 1 1 5]]
h = [1 0 1 1 1]

note h(1) = 1 not 2 even though there are 2 occurrences of 1 in u

I know the following loop will count properly

for i=1:n
h(u(i)) = h(u(i)) +1;
end

How can I code this a a vector operation without a loop?

Answer 1

[Azzi Abdelmalek]

u = [3 4 1 1 5]
accumarray(u',[1:numel(u)]',[],@(x) numel(x))

[Jan]

The additional square brackets are not needed and waste time only:

[1:numel(u)]'  ==> faster: (1:numel(u))'