MATLAB Answers

Evaluating 3D Function

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Brian Lee
Brian Lee on 2 Feb 2012
Edited: Matt J on 26 Sep 2013
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

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Answers (2)

Sean de Wolski
Sean de Wolski on 2 Feb 2012
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

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Benjamin Schwabe
Benjamin Schwabe on 2 Feb 2012
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

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Sean de Wolski
Sean de Wolski on 2 Feb 2012
Not true, meshgrid and ndgrid both support higher dimensions.
Walter Roberson
Walter Roberson on 2 Feb 2012
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
Sean de Wolski on 2 Feb 2012
Interesting, I stand corrected. I could've sworn it did the same thing as ndgrid just with x/y swapped with r/c.

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