Is there a better vectorization technique?

I am trying to see if there are other ways of coding this code sample more efficiently. Here, y is an 1xM matrix, (say, 1x1000), and z is an NxM matrix, (say, 5x1000).

   mean(ones(N,1)*y.^3 .* z,2)

This code works fine, but I worry of N increases a lot, that the:

   ones(N,1)*y.^3

might get too wasteful and make everything slow down.

Thoughts?

Answer 1

[John D'Errico]

Its not THAT terrible for a matrix that small. However, you can gain from the use of bsxfun in many instances. Here, the matrices are simply too small to really gain anything.

>> N = 5;M =1000;
>> y = rand(1,M);
>> z = rand(N,M);
>> mean(ones(N,1)*y.^3 .* z,2)
ans =
      0.12412
      0.11669
      0.12102
      0.11976
      0.12196
>> mean(bsxfun(@times,y.^3,z),2)
ans =
      0.12412
      0.11669
      0.12102
      0.11976
      0.12196
>> z*y.'.^3/M
ans =
      0.12412
      0.11669
      0.12102
      0.11976
      0.12196

As you can see, all three solutions return the same result. All are equally valid.

Now I'll compare the times required.

>> timeit(@() mean(ones(N,1)*y.^3 .* z,2))
ans =
   0.00023018

>> timeit(@() mean(bsxfun(@times,y.^3,z),2))
ans =
   0.00026829

>> timeit(@() z*y.'.^3/M)
ans =
   0.00016594

As I said, you don't gain much. In fact, bsxfun does not gain at all, and is a bit slower. But you can gain a bit, if you re-write the expression into the third form I've posed. Not much, but a bit.

[Kalamaya]

I thought about using bsxfun. How is the timing if the N was very large, say, 2000 or something like that? (timeit is your own function?)

[Jan]

FEX: timeit

Answer 2

[Jan]

The power() operator is more expensive than the matrix multiplication. Therefore an explicit multiplication saves time:

M = 1000;
N = 5;
y = rand(1, M);
z = rand(N, M);

tic; for i=1:100, a = mean(ones(N,1) * y .^ 3 .* z, 2); end; toc
% Elapsed time is 0.036668 seconds.

tic; for i=1:100, a = z * y.' .^ 3 / M; end; toc
% Elapsed time is 0.026818 seconds.

tic; for i=1:100, a = z * (y .* y .* y)' / M; end; toc
% Elapsed time is 0.001327 seconds.

[EDITED] If the resulting array is very large, a multiplication is faster than a division, but the result can differ due to rounding:

a = z * (y .* y .* y)' * (1 / M);

For the small [5x1] array in the example this does not matter.