Using bsxfun instead of arrayfun for repeating function

I am using a function that produces row vectors based on the RNG. Now I would like to repeat that function and put the rows into a matrix. I am currently using arrayfun, but that seems inneficient to me. Is there a way to use bsxfun to do this?(I included an example code to show what I mean)

Jos van den Berg

u=4;
v=5;
h=6;

cell2mat(arrayfun(@(i) myfun(v,u,h) , 1:3, 'UniformOutput', false )')

function y = myfun(v,u,h)
 y = [u*rand,v*rand,h*rand];
end

Answer 1

[Guillaume]

bsxfun serves a completely different purpose to arrayfun. One rarely replaces the other. Note that as of R2016b, for most operations, bsxfun is not needed anymore.

In your particular example, the whole output could be obtained simply with:

y = [u * rand(3, 1), v * rand(3, 1), h * rand(3, 1)];

or if you really wanted to use bsxfun:

y = bsxfun(@times, [u, v, h], rand(3, 3));

and as said, as of R2016b, the bsxfun can be replaced by implicit expansion:

y = [u, v, h] .* rand(3, 3);

[JJP van den Berg]

Thank you for your answer. I guess I mix up the uses of these funtions. I get that I can get the ouptut in a different way, however the function I am using is more complicate. Maybe I should have posed the question differently. I would like to repeat a function the fastest way and store the output which is a row vector in a matrix. My understanding is that loops are not very fast, so I went looking for a different way and got to the arrayfun and bsxfun functions. But looking at yout answer I can conclude that bsxfun is not the way to do this fast right?

[Guillaume]

bsxfun is only useful, if your unknown function takes exactly two inputs (scalar, vectors, or matrices) and returns one input. It's certainly not designed to repeatedly call a function.

For that, indeed arrayfun or a loop is the way to go. I tend to prefer arrayfun as it makes it clear what is being repeated and what it is repeated over, but an explicit loop can actually be faster as it avoids the cost of calling an anonymous function.

[JJP van den Berg]

Not really sure how I got to this misconception of bsxfun. Thanks a lot for the explanation!