Question

Suppose I have a matrix such that each row is a standard basis vector, i.e. each row contains exactly one 1, the other columns being 0.

Is there a convenient way to create such a matrix (i.e. given a vector of positions of where the ones are in each row)?

Also, is there a way I should represent such a matrix so that multiplications with it can be done more efficiently in octave?

Answer 1

Suppose you want a 3×3 matrix with the ones in columns 3, 1, and 2 respectively:

> pos = [3,1,2];
> x = eye(3)(pos,:);

will give you a matrix with 9 elements, most zero, with the ones in the desired places. You can save memory by using a sparse representation: sparse_x = sparse(x);. But the following test on my machine implies that the natural form multiplies faster:

> N = 10000;
> s = rand(N,N);
> x = eye(N)(randperm(N),:);
> sx = sparse(x);
> t = cputime(); ss = s*x; cputime()-t
ans = 0.41124
> t = cputime(); ss2 = s*sx; cputime()-t
ans = 1.0313

This was Octave 3.4 on a Core i7, YMMV.

Looking at whos it appears that Octave is doing something clever with x:

> whos
Variables in the current scope:

  Attr Name        Size                     Bytes  Class
  ==== ====        ====                     =====  ===== 
       N           1x1                          8  double
       s       10000x10000              800000000  double
       ss      10000x10000              800000000  double
       ss2     10000x10000              800000000  double
       sx      10000x10000                 160004  double
       x       10000x10000                  40000  double  <---SMALLER THAN s!

If it knows x is special, maybe it’s already taking advantage of speedups in the multiplication.