Counting unique values across the columns of a matrix
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How can I store indices of columns in a matrix containing more than 3 unique values? for example if: X =
8 2 1 11 0
8 10 11 3 4
9 14 6 1 4
4 3 15 11 0
I want Y=[2 3]
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Cedric
2015년 7월 24일
편집: Cedric
2015년 7월 24일
UPDATED as developed in the comment below
y = find( arrayfun( @(c) numel( unique( A(:, c) )), 1:size( A, 2 )) > 3 ) ;
FORMER
Assuming that the matrix is stored in variable A, here is an ugly one-liner:
>> find( arrayfun( @(c) nnz( accumarray( 1+A(:,c), 1 )) > 3, 1:size( A, 2) ))
ans =
2 3
But your condition "more than 3 unique values" with 4 rows, means "all unique per column". Is it a small study case and then you will need fewer unique elements per column, or will this always be "all unique"?
Here is another ugly solution:
>> find(sum(accumarray([1+A(:), reshape(bsxfun(@times, 1:size(A,2), ones(size(A))), [], 1)], 1)>0, 1)>3)
ans =
2 3
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Cedric
2015년 7월 24일
편집: Cedric
2015년 7월 24일
Hi Matt, I cannot see your attachment, but here is a better option which doesn't assume that elements of your array are integers starting at 0:
y = find( arrayfun( @(c) numel( unique( A(:,c) )), 1:size( A, 2 )) > 3 ) ;
PS: I suspect that the solution based on ACCUMARRAY is faster than the one based on UNIQUE, so if time is an issue, it may be worth adapting the former solution, working on a suitable offset for replacing the 1 in 1+A...
추가 답변 (1개)
Andrei Bobrov
2015년 7월 24일
[~,ii] = mode(X);
out = find(size(X,1) - ii >= 3);
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Andrei Bobrov
2015년 7월 24일
편집: Andrei Bobrov
2015년 7월 24일
>> X = [
8 2 1 11 0
8 10 11 3 4
9 14 6 1 4
4 3 15 11 0];
>> [~,ii] = mode(X);
>> out = find(size(X,1) - ii >= 3)
out =
2 3
>> out = find(size(X,1) - ii > 3)
out = [](1x0) % solution in Octave (now I can't use MATLAB)
>>
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