Remove the rows/columns with single values of a matrix

Question

Dani Tormo 2012-11-28

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Hello,

I have a matrix quite big (A < 23 x 250000 >). I want to evaluate the first row A(1,:) and eliminate the columns where there are values only repeated once. The number in this first row is an integer and always is increasing.

Lets say:

 B = [1    1    3    5    5    5    7    9    9
      0.1  0.5  0.2  0.4  0.3  0.9  0.1  0.6  0.5]

The result should be:

 B = [1    1    5    5    5    9    9
      0.1  0.5  0.4  0.3  0.9  0.6  0.5]

I know how to do it with for and if statements but I think it won't be efficient because the size of the matrix.

Many thanks!

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Answer 1

Dani Tormo 2012-11-28

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This is the code I have and it works, but maybe it is not efficient:

 for i = 2:length(A) - 1;
    if A(1,i) ~= A(1,i-1) && A(1,i) ~= A(1,i+1)
        A = A(:,[1:i-1 i+1:end]);
        i = i - 1;
    end
 end

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Dani Tormo 2012-12-3

编辑：Dani Tormo 2012-12-3

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Yes, finally I had to do it with the while expression to avoid the for fail. The rush isn't good ever.

 finished = 0; i = 2;
 while finished == 0
    if A(1,i) ~= A(1,i-1) && A(1,i) ~= A(1,i+1)
        simulation_data = simulation_data(:,[1:i-1 i+1:end]);
    elseif i == length(A)-1
        finished = 1;
    else
        i = i + 1;
    end
 end

Regards.

Jan 2012-12-3

编辑：Jan 2012-12-3

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Collect the indices to be removed at first:

toDelete = false(size(A, 2));
for i = 2:size(A, 2) - 1
  if A(1,i) ~= A(1,i-1) && A(1,i) ~= A(1,i+1)
     toDelete(ii) = true;
  end
end
A(:, toDelete) = [];

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Answer 2

per isakson 2012-11-28

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https://ww2.mathworks.cn/matlabcentral/answers/55024-remove-the-rows-columns-with-single-values-of-a-matrix#answer_66770

编辑：per isakson 2012-11-28

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A start:

    B = [1    1    3    5    5    5    7    9    9
          0.1  0.5  0.2  0.4  0.3  0.9  0.1  0.6  0.5];
    [ uniqueB, ix ] = unique( B(1,:) );
    sB = B(1,:);
    sB( ix ) = [];
    singleB = setdiff( uniqueB, sB );
    [ ~, ixm ] = ismember(  singleB, B(1,:) );
    B( :, ixm ) = []

prints

    B =
        1.0000    1.0000    5.0000    5.0000    5.0000    9.0000    9.0000
        0.1000    0.5000    0.4000    0.3000    0.9000    0.6000    0.5000
    >>

your turn.

I'm wouldn't be surprise if the for-loop is faster :)