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

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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

Risposta accettata

Birdman
Birdman il 23 Ott 2017
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.
  2 Commenti
Jan
Jan il 23 Ott 2017
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
Soabon il 24 Ott 2017
Yes, with the correction this is what I needed (and it works with MATLAB 2015b). Thanx!

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Più risposte (2)

Jan
Jan il 23 Ott 2017
Modificato: Jan il 24 Ott 2017
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;
  4 Commenti
Soabon
Soabon il 26 Ott 2017
Modificato: Soabon il 26 Ott 2017
The size of RT is 51 x 50 and the length of krit_out1 is 50. Does that make a difference for the solution?

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Ray
Ray il 23 Ott 2017
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
  4 Commenti
Jan
Jan il 24 Ott 2017
Modificato: Jan il 24 Ott 2017
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
Soabon il 26 Ott 2017
You're right, I applied the code a second time, now it works. I don't know why there was the error message before.

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