操作矩阵的所有行，不循环，八度

Question

Suppose I have a matrix A and I want to obtain the following:

for i=1:m
  A(i,:) = something which depends on i;
endfor

Is there a way to obtain that without the loop?

Added: Ok, I've understood I have to be more specific.
I have two matrices B and C (all the matrices we are considering have m rows).
I want to record in the i -th row of A the product of the polynomials written in the i -th rows of B and C (so using the loop I would call the conv function). Any ideas?

Answer 1

That is a very very general question, and not possible to answer with more details. Mainly on what i will be involved with. Suppose the following

for i = 1:m
  A(i,:) += i;
endfor

It could be written with the much more efficient:

A .+ (1:m)'

Just compare:

octave> n = 1000;
octave> A = B = rand (n);
octave> tic; for i = 1:n, B(i,:) += i; endfor; toc
Elapsed time is 0.051 seconds.
octave> tic; C = A.+ (1:n)'; toc
Elapsed time is 0.01 seconds.
octave> isequal (C, B)
ans =  1

If you have a very old version of octave, you can instead do bsxfun (@plus, A, (i:m)') .

However, if i on the right side of the expression will be used for indexing some other variable, then solution would be different. Maybe, the solution is cumsum , or some other of cumfoo function.

Your question is basically, "how do I vectorize code?", which is a really large subject, without telling us what you're trying to vectorize.

Answer 2

I don't think it's possible to do this without a for loop as conv only accepts vector inputs, but I might be wrong. I can't see a way of using either bsxfun or arrayfun in conjunction with conv and matrix inputs. I might be wrong though... I stand to be corrected.