Vectorizing inside a loop

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Avery
Avery el 20 de Mzo. de 2013
I am trying to implement something like this
for i=1:n
matrix(i:n:end,4:6)=y(i,:)-matrix(i,1:3);
end
I was under the impression that vectorizing would allow me to eliminate the need for another loop but I receive 'Subscripted assignment dimension mismatch' when I attempt to evaluate this. I believe it is because it is not assigning it on a per row basis as I had hoped.
Is it at all possible to accomplish this without adding more loops?
  3 comentarios
Avery
Avery el 20 de Mzo. de 2013
y is an n by 3 matrix. n changes, hence it being a variable, as do the size of y and matrix. I understand that the left is a matrix and the right is a vector, but my understanding of vectorization leads me to believe that I can replace a loop in this way.
ChristianW
ChristianW el 20 de Mzo. de 2013
Show your working double loop.

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Respuestas (1)

Babak
Babak el 20 de Mzo. de 2013
The problem with
matrix(i:n:end,4:6)=y(i,:)-matrix(i,1:3);
is that the first index i:n:end is a vector and does not correspond to 1 element. On the other hand, on the righe side, the index i is a sinle element which chooses the row with index i.
You need to change i:n:end to a proper single index accordign to what you want to write to. If you give a numerical example, maybe I can figure what you really want to do.

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