Building a matrix in a faster way

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Azza
Azza el 16 de Oct. de 2012
Hi,
I am trying to build a matrix by giving each array in the matrix the same value in its first column. The value is [0;0;1]. My code look something like this:
yv = 1:-1:-1;
xv = -1:1:1;
for Y = 1:length(yv)
for X = 1:length(xv)
M(:,1,X,Y) = [0;0;1];
end
end
I was wondering if there is more efficient way to give the arrays for length (yv) and (xv) the value [0;0;1] instantly without using the for loop. My matrix in original is much larger than this and I need to make the code as faster to execute the data as possible.
Highly appreciate any help with this.
Best wishes
AA

Respuesta aceptada

Matt J
Matt J el 16 de Oct. de 2012
d=[0;0;1];
M=d(:,1,ones(1,length(xv)), ones(1,length(yv)))
  2 comentarios
Walter Roberson
Walter Roberson el 16 de Oct. de 2012
Which can also be written as
M = repmat(d, [1, 1, length(xv), length(yv)]);
Matt J
Matt J el 16 de Oct. de 2012
Yes, although repmat does use mcode containing loops, and therefore can be slow.

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Más respuestas (1)

Azza
Azza el 17 de Oct. de 2012
Many thanks Matt and Walter for your help. Both your answers are very valuable. The thing with my code is that I need to keep a counter in each line. For example the code with the for loop should look something like this:
for Y = 1:length(yv)
for X = 1:length(xv)
counter = 1;
M(:,counter,X,Y)= [0;0;1];
counter = counter;
M(:,counter,X,Y) = A*Rflip*M(:,1,X,Y)+B;
end
end
Thus, the counter should change from value 1 to 2 accordingly.
I have also replicated the matrix for A, Rflip and B in order to accomodate the M value for the length of arrays of (xv) and (yv) similar to your methods. The original sizes of matrices A and Rflip were 3*3 for each element. So I managed to replicate the matrix to [3 3 3 3] for (xy) and (xv) While for B was 3*1 and I made it into [3 1 3 3]. When I tried to execute the line with the replicated matrices for A, Rflip and B {while excluding the counter} I got this error message:
??? Error using ==> mtimes Input arguments must be 2-D.
So would you kindly help me in giving the length of arrays for (xv) and (xy) the same value of M without using the lengthy for loop method while including the counter?
Best wishes
AA
  1 comentario
Matt J
Matt J el 17 de Oct. de 2012
Perhaps. But first Accept-click the answer we gave you and then start a new post for your new question.

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