'' Attempted to access sorted_evalue(1); index out of bounds because numel(sorted_evalue)=0. '' how to remove this???

 %calculate eigen vector & eigen value of cm = evector & evalue
    evector = dfdb * rvector;
    evalue = diag( rvalue );
    %clear rvector rvalue
    disp( 'Calculated Eigen Vectors & Eigen Values of CM' );
    [sorted_evalue, index] = sort( evalue );            %sorted in acending order
    sorted_evalue = flipud( sorted_evalue );            %rearranged in decending order
    index = flipud( index );                            %rearranged corresponding indies also
    %Now rearrange eigenvectors in the order of rearranged eigen values
    evector( :, 1:nof ) = evector ( :, index );
    smallest_evalue = (1/100) * sorted_evalue(0);            % to obtain the effective eigen values
    for i = 1:nof
        if (sorted_evalue(i) < smallest_evalue )
            break;
        end
        evalue = [ ];
    end
    index = i+1;

Answer 1

[Walter Roberson]

%calculate eigen vector & eigen value of cm = evector & evalue
    if isempty(dfdb) || isempty(rvector)
      error('Need a non-empty dfdb and rvector');
    end
    evector = dfdb * rvector;
    evalue = diag( rvalue );
    %clear rvector rvalue
    disp( 'Calculated Eigen Vectors & Eigen Values of CM' );
    [sorted_evalue, index] = sort( evalue, 'descending' );
    %Now rearrange eigenvectors in the order of rearranged eigen values
    if nof ~= length(index)
      error( sprintf('nof found to be %d, needs to match number of indices = %d', nof, length(index)) );
    end
    evector( :, 1:nof ) = evector ( :, index );

      %sorted-evalue are in descending order. Are you *sure* you want to take a fraction
      %of the largest of them and name that as "smallest_evalue" ??

      smallest_evalue = (1/100) * sorted_evalue(1);            % to obtain the effective eigen values

      %What is the point of this loop? If it is to find the _second_ location where an eigen value
      %is less than 1/100 of the maximum value, then there would be much more straightfoward
      %ways of doing this

      for i = 1:nof
          if (sorted_evalue(i) < smallest_evalue )
              break;
          end
          %and why the heck is evalue left alone if the largest eigenvalue is negative
          %but cleared otherwise? Because the only way that
          % sorted_evalue(1) < sorted_evalue(1)/100 immediately is if sorted_evalue(1) < 0

          evalue = [ ];
      end

      %and if we reached the end of the loop without using the "break" then is it
      %truly okay that index will point past all of the values??

      index = i+1;

Answer 2

[Guillaume]

It would be helpful to know which line is causing the error. It should be part of the error message.

With the code you've posted, the only line I can see could generate this error is the if test within the for loop. The thing is, there's no way your code could reach that line because another error will occur before that.

Anyway, the error message is very clear, sorted_evalue has 0 elements. It's empty. If it's empty that means that evalue and in turn rvalue are empty. Only you can tell why they are empty.

But as I said, I don't see how your code can even generate the error, since it will fail on the line.

smallest_evalue = ... sorted_evalue(0)

This will throw the error "Subscript indices must either be real positive integers or logicals." since indexing in matlab starts at 1.

[Guillaume]

By the way, learn to use the debugger. Step through your program and see how the variables evolve. You'll quickly see where it goes wrong.

You'll also find out that your loop probably does not do what you want (it does nothing, and if it did anything it would empty the whole evalue.)