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

1 次查看（过去 30 天）
Soabon
Soabon 2017-10-23
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

请先登录，再回答此问题。

采纳的回答

Birdman
Birdman 2017-10-23
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 个评论
Jan
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.

请先登录，再进行评论。

更多回答（2 个）

Jan
Jan 2017-10-23
编辑：Jan 2017-10-24
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 个评论
显示 2更早的评论
Soabon
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?

请先登录，再进行评论。

Ray
Ray 2017-10-23
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 个评论
显示 2更早的评论
Jan
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
Soabon 2017-10-26
You're right, I applied the code a second time, now it works. I don't know why there was the error message before.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Logical 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by