Counting occurrences of a pointer

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Alvin Stroyny
Alvin Stroyny 2013-8-5
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?

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采纳的回答

Azzi Abdelmalek
Azzi Abdelmalek 2013-8-5
编辑：Azzi Abdelmalek 2013-8-5
u = [3 4 1 1 5]
accumarray(u',[1:numel(u)]',[],@(x) numel(x))
  1 个评论
Jan
Jan 2013-8-5
The additional square brackets are not needed and waste time only:
[1:numel(u)]' ==> faster: (1:numel(u))'

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