most frequent value in a 3D matrix other than 0

Dear all

I have got a 3D matrix, say A ( 100 by 100 by 20) containing elements of '0' , '1' , '2', '3', '4'.

I would like to know which elements happens the most frequent, except for '0'. My idea is to reshape the matrix to a vector using

A=reshape(A,100*100*20,1);
numbers = unique(A);   % sorted occurrence 
count=hist(A,numbers);
pointer = find(count == max(count(2:end))); % by using (2:end), I excluded the effect by '0'
value = numbers(pointer);

However, this seems redundant, any one have a better way to do that? Say, dealing with the 3D matrix straight. My 3D matrix will be very large....

Best

Yuan Chen

[aiman saad]

hi every body , there is more simple built-in function , i just found that satisfy the title requirement : search the help for ( mode , M = mode(A,dim) ) Most frequent values in array good luck for all

Answer 1

[Azzi Abdelmalek]

A=randi([0 4],100,100,20); %Example
[freq,number]=max(histc(A(:),1:4))

[Yuan chen]

Thanks a lot dear Azzi, this helps a lot :)

Best

Yuan

Answer 2

[Cedric]

Here is an example..

 >> A = randi(5,[3,4,2]) - 1
 A(:,:,1) =
     0     3     1     4
     2     0     2     0
     3     3     4     1

 A(:,:,2) =
     3     4     4     0
     0     3     2     2
     1     0     4     4

 >> counts = accumarray( A(:)+1, ones(numel(A),1) )     % or HISTC
 counts =
     6
     3
     4
     5
     6

 >> mostFrequent = find( counts == max(counts) ) - 1
 mostFrequent =
     0
     4

EDIT: note that if you want to avoid taking 0's, you can do

 >> mostFrequent = find( counts(2:end) == max(counts(2:end)) )
 mostFrequent =
     4

I kept zeros to show that several elements can be the most frequent, so whether you use Azzi's HISTC solution or my ACCUMARRAY approach, you shouldn't use the second output argument of MAX to get the most frequent element, otherwise you'll only get the first of them (if there are multiple).

[Yuan chen]

Thanks very much Cedirc :)

[Cedric]

Whether you use ACCUMARRAY or HISTC, it would actually be more efficient to create a vector of non-zero values first..

 >> nz = A(A~=0) ;
 >> counts = accumarray( nz, ones(size(nz)) ) ;
 >> mostFrequent = find( counts == max(counts) )
 mostFrequent =
     4

[Yuan chen]

Hi Cedric

That's very thoughtful, I was just thinking about that.!

Best