Counting unique values across the columns of a matrix

Question

Matt Talebi 2015-7-24

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https://ww2.mathworks.cn/matlabcentral/answers/230994-counting-unique-values-across-the-columns-of-a-matrix

编辑： Andrei Bobrov 2015-7-24

How can I store indices of columns in a matrix containing more than 3 unique values? for example if: X =

   2     1    11   0
  10    11     3   4
  14     6     1   4
   3    15    11   0

I want Y=[2 3]

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Answer 1

Cedric 2015-7-24

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编辑：Cedric 2015-7-24

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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 2015-7-24

编辑：Cedric 2015-7-24

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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...

Matt Talebi 2015-7-24

This one works just fine! Thank you so much for your time.

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Answer 2

Andrei Bobrov 2015-7-24

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[~,ii] = mode(X);
out = find(size(X,1) - ii >= 3);

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Matt Talebi 2015-7-24

Thanks Andrei for your attention. I've modified your codes by replacing ii>=3 with ii>3 (which is what I want) but still they don't return correct answer.

Andrei Bobrov 2015-7-24

编辑：Andrei Bobrov 2015-7-24

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>> 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)
>>