Evaluating 3D Function

I have a 3-D function f(x,y,z) which I want to evaluate at a discrete set of points (i.e. create an array with values of f).

My x,y,z variables are not equally spaced. What is the best way to define f?

For a 1-D function, I would normally use array operators:

x=linspace(1,5,100); f=x.^2;

For speed reasons I would like to avoid for loops.

But for higher dimensions this stops working:

x=linspace(1,5,100); y=linspace(1,5,10); f=x.^2+y.^2; <----does not work because of diff array sizes

Answer 1

[Sean de Wolski]

I would use bsxfun() to generate the grid of points in 3d. If this approach wouldn;t work, you could always use ndgrid() and actually generate the grid matrices.

Example with bsxfun:

x = [0 .3 5 100];
y = linspace(1,10,20);
z = logspace(1,2,3);

f = bsxfun(@plus,bsxfun(@plus,x',y),reshape(z,1,1,numel(z))) %just adding - nothing fancy

Answer 2

[Benjamin Schwabe]

Hi,

another easy way to do that, is using the meshgrid command.

x = linspace(1,5,100); y=linspace(1,5,10); [X,Y] = meshgrid(x,y); F = X.^2+Y.^2;

If you want to go to even higher dimensions, you will require multidimensional arrays and for loops.

Benjamin

[Sean de Wolski]

Not true, meshgrid and ndgrid both support higher dimensions.

[Walter Roberson]

Sean: http://www.mathworks.com/help/techdoc/ref/meshgrid.html

"The meshgrid function is similar to ndgrid, however meshgrid is restricted to 2-D and 3-D while ndgrid supports 1-D to N-D."

[Sean de Wolski]

Interesting, I stand corrected. I could've sworn it did the same thing as ndgrid just with x/y swapped with r/c.