Hello there, I have a short question, as for some reason a for loop doesn't function.

RT is a matrix and krit_out1 is a vector. There is no error message, but the loop is stuck in the first row of the matrix and I don't find the reason.. Can someone help me?

for i = 1:50
    RT(RT(:,i)> krit_out1(i)) = NaN;
end

[Stephen23]

Loops are not required to solve this. See Jan Simon's answer for a simple and efficient solution.

Answer 1

[Birdman]

RT=ones(50,50);krit_out1=zeros(50,1);
for i=1:50
    for j=1:50
        if(RT(j,i)>krit_out1(j))
            RT(j,i)=NaN;
        end
    end
end

Try this.

[Jan]

krit_out1( i ) instead of j.

The vectorization of such loops is not only nice and processed efficiently, but without indices, there are less chances for typos.

[Soabon]

Yes, with the correction this is what I needed (and it works with MATLAB 2015b). Thanx!

Answer 2

[Jan]

Or without a loop:

RT(RT > krit_out1(:).') = NaN;   % >= R2016b: Auto-Expanding

For older Matlab versions:

index     = bsxfun(@gt, RT, krit_out1(:).');   % [EDITED]
RT(index) = NaN;

[Soabon]

Thank you, this looks really nice, but I use an older version 2015b and get an error message :-/

"Error using bsxfun Non-singleton dimensions of the two input arrays must match each other."

[Jan]

See [EDITED]. It is easier to write a matching answer, if you provide the sizes of the used arrays.

[Soabon]

The size of RT is 51 x 50 and the length of krit_out1 is 50. Does that make a difference for the solution?

[Soabon]

Now this solution also worked well.

Answer 3

[Ray]

Try the following. It looks like you intend to operate on the ith column using the ith element of a vector called krit_out1:

for i = 1:50
    RT(RT(:,i)> krit_out1(i) ,i) = NaN;
end

[Jan]

+1. You hit the point.

[Soabon]

Thank you, this looks like a good solution, unfortunately I get an error message applying this. "Subscript indices must either be real positive integers or logicals."

[Jan]

Ray's code uses only the same indices as the code you have posted. Therefore this piece of code cannot produce the error message, except if you are using a different loop counter, but insert "i".

This method is better than using another loop.

[Soabon]

You're right, I applied the code a second time, now it works. I don't know why there was the error message before.