Error using horzcat in a for loop

Question

Sebastian 2014-1-21

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/113158-error-using-horzcat-in-a-for-loop

评论： Sebastian 2014-1-21

Hi everyone, I am testing a for loop and it is showing me some error in the horzcat, I do not know why it does. Can anyone explain me why it happens, and help me to fix my code. Thank you

p=1;

[row,col] = size(A);

for i = 1:row

    for j=1:col
        if(A(i,j)>0)
            z(p,:)=[j,i];
            p=p+1;
        end
    end

end

B=[(1:p-1)',z]

This is the error:

Error in ==> testloop at 11

B=[(1:p-1)',z]

??? Error using ==> horzcat

CAT arguments dimensions are not consistent.

Error in ==> testloop at 11

B=[(1:p-1)',z];

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

kjetil87 2014-1-21

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/113158-error-using-horzcat-in-a-for-loop#answer_121673

编辑：kjetil87 2014-1-21

在 MATLAB Online 中打开

horzcat concatinates arrays in horizontal direction i.e x = [a , b] is horzcat. when you call B=[(1:p-1)' ,z] your are calling horzcat (there is also a version called vertcat).

The problem is that (1:p-1)' and z does not have the same amount of rows. Which is strange, because your code looks correct. perhaps z allready existed before you ran the code? try adding

clear z;

before your loop.

or

B=[(1:p-1)' , z(1:p-1,:)];

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Sebastian 2014-1-21

It worked, Thank you very much I undestand now why it happens

请先登录，再进行评论。

Answer 2

Amit 2014-1-21

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/113158-error-using-horzcat-in-a-for-loop#answer_121675

在 MATLAB Online 中打开

I think I figured it out.

This error must be occurring not everytime. The thing that if you run this few time, the size of z can change. But as you did not cleared the z from previous run, in new case (where size of z should be lower), the new value only replace in the previous locations and does not update the size of z. Try the code below. this will work.

p=1;
[row,col] = size(A);
clear z;    % This is needed
for i = 1:row
    for j=1:col
        if(A(i,j)>0)
            z(p,:)=[j,i];
            p=p+1;
        end
    end
end
B=[(1:p-1)',z]

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

Sebastian 2014-1-21

Both answers were correct, and since you told me this I remember that everytime I open the code the first matrix works, but the second does not. It should be that, Thank you

请先登录，再进行评论。