most frequent value in a 3D matrix other than 0
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Yuan chen
am 19 Dez. 2013
Kommentiert: aiman saad
am 22 Jun. 2016
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
1 Kommentar
aiman saad
am 22 Jun. 2016
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
Akzeptierte Antwort
Azzi Abdelmalek
am 19 Dez. 2013
Bearbeitet: Azzi Abdelmalek
am 19 Dez. 2013
A=randi([0 4],100,100,20); %Example
[freq,number]=max(histc(A(:),1:4))
Weitere Antworten (1)
Cedric
am 19 Dez. 2013
Bearbeitet: Cedric
am 19 Dez. 2013
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).
3 Kommentare
Cedric
am 19 Dez. 2013
Bearbeitet: Cedric
am 19 Dez. 2013
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
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