Error using horzcat in a for loop

3 visualizaciones (últimos 30 días)
Sebastian
Sebastian el 21 de En. de 2014
Comentada: Sebastian el 21 de En. de 2014
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];

Respuesta aceptada

kjetil87
kjetil87 el 21 de En. de 2014
Editada: kjetil87 el 21 de En. de 2014
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 comentario
Sebastian
Sebastian el 21 de En. de 2014
It worked, Thank you very much I undestand now why it happens

Iniciar sesión para comentar.

Más respuestas (1)

Amit
Amit el 21 de En. de 2014
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 comentario
Sebastian
Sebastian el 21 de En. de 2014
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

Iniciar sesión para comentar.

Categorías

Más información sobre Loops and Conditional Statements en Help Center y File Exchange.

Community Treasure Hunt

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

Start Hunting!

Translated by