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

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Soabon 2017-10-23

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评论： Soabon 2017-10-26

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

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Stephen23 2017-10-24

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

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Birdman 2017-10-23

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在 MATLAB Online 中打开

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.

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Jan 2017-10-23

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 2017-10-24

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

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Jan 2017-10-23

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编辑：Jan 2017-10-24

在 MATLAB Online 中打开

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;

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Soabon 2017-10-26

编辑：Soabon 2017-10-26

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 2017-10-26

Now this solution also worked well.

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Ray 2017-10-23

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在 MATLAB Online 中打开

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

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Jan 2017-10-24

编辑：Jan 2017-10-24

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 2017-10-26