Finding the repetition of each element of a matrix

I have a matrix named A with size N×3. For example,

A = [ 1     2     3;
      1     4     5;
      5     6     7;
      1     3     4;
      5     6     7;
      1     4     7;
      4     3     2;
      7     8     9]

In this example, The elements of A are:

    B = [1 2 3 4 5 6 7 8 9]

How can I find the repetition of each element of B in A? Actually I am looking for a a vector array like C:

    C = [4 2 3 4 3 2 4 1 1]

Here is what I did:

    for i=1:numel(B)
        C(i) = numel(find(ismember(A, i)==1));
    end

I would appreciate it if someone could help me how I can find the repetition of each element of A in a better way and avoid the for-loop.

Answer 1

[Matt J]

Come to think of it, this might be better than what I first proposed (and have now deleted)

C=accumarray(A(:),1); 
 C=C(B);

Solutions based on HISTC only work if the B(i) are sequential and increment by 1.

[Image Analyst]

What is B? I assume it's unique(A).

By the way, the histc() method works on integer lists if they don't increment by 1. I tried it before I posted - changed the 8 to 18. You just get zeros for the counts on the "in-between" missing numbers.

If you have gaps in the integers but don't want gaps in the histogram, you can use B=unique(A:)) and it works just fine:

A = [ 1     2     3;
  1     4     5;
  5     6     7;
  1     3     4;
  5     6     7;
  1     4     7;
  4     3     2;
  7     18     9]
B = unique(A)
counts = histc(A(:), B)

You won't have bins with zero counts in them. Just be aware that the bin centers are not sequential. For example (using the above example) bin 7 counts values of 7, bin 8 is for 9, and bin 9 is for 18.

If the numbers aren't integers, that complicates things greatly and the FAQ must be understood: http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F

[Matt J]

What is B? I assume it's unique(A).

I don't know what the OP intended, but conceivably B can be any requested subset of elements in A, and in any order. So, for example if B=[9,1],

    >> C=accumarray(A(:),1); C=C(B)

    C =

         1
         3

I guess you could also do

    >> C=histc(A(:), 1:max(B)); C=C(B)

    C =

         1
         3

So, fine. There is a histc-based option. I wonder, though, whether accumarray isn't more efficient when B isn't of the form 1:N.

Answer 2

[Image Analyst]

Try this:

edges = min(A(:)) : max(A(:))
counts = histc(A(:), edges)