[SciPy-User] Creating meshgrid from meshgrid

[Florian Lindner]

Hello,

yes, the subject sound a bit weird...

I have two arrays of size N (let's say 2 arrays of length 4) that I combine using np.meshgrid

xxA, yyA = np.meshgrid(xA, yA)
xxB, yyB = np.meshgrid(xB, yB)

which gives me two meshes

xx.shape = yy.shape = (4,4)
which represent a N-dimensional mesh with 16 elements.

Now I want to evaluate a function f on every possible pair of N-dimensional points in the grid, resulting in a 16 x 16
matrix:

in a flattened notation, pA = (xxA, yyA)

f(pA[1]-pB[1]) f(pA[1]-pB[2]) f(pA[1]-pB[3]) ...
f(pA[2]-pB[1]) f(pA[2]-pB[2]) f(pA[2]-pB[3]) ...
f(pA[3]-pB[1]) f(pA[3]-pB[2]) f(pA[3]-pB[3]) ...
.
.
.


Let's say xA = yA = [1,2,3] and xB = yB = [10,20,30]

that gives me a mesh A:

(1,3) (2,3) (3,3)
(1,2) (2,2) (3,2)
(1,1) (2,1) (3,1)

and a mesh B alike.

My result matrix now should be of size 9 x 9:

f( (1,3), (10,30) ) f( (2,3), (20,30) ) f( (3,3), (30, 30) )
f( (1,2), (10,20) ) f( (2,2), (20,20) ) f( (3,2), (30, 20) )
...


f always takes two N-dimensional vectors and returns a scalar.

I hope I was able to explain what I want to achieve.

What is the best way to do that in numpy/scipy?

As long as the meshes itself are 1-d I did it like that:

mgrid = np.meshgrid([1,2,3], [10,20,30])
A = f( np.abs(mgrid[0] - mgrid[1]) )

Thanks,
Florian