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
il 24 Lug 2015
Modificato: Cedric
il 24 Lug 2015
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
3 Commenti
Cedric
il 24 Lug 2015
Modificato: Cedric
il 24 Lug 2015
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...
Più risposte (1)
Andrei Bobrov
il 24 Lug 2015
[~,ii] = mode(X);
out = find(size(X,1) - ii >= 3);
2 Commenti
Andrei Bobrov
il 24 Lug 2015
Modificato: Andrei Bobrov
il 24 Lug 2015
>> 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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