How to convert a point in meshgrid to vector coordinates?

I have a meshgrid [X,Y,Z] = meshgrid(x,y,z). Let's say the meshgrid is size K*K*K and each of X,Y,Z are individually also of this size. How do I go from a point in the meshgrid to coordinates of X/Y/Z? Sorry if that doesn't make any sense but maybe that is why I am confused!

[KSSV]

What do you mean by that? Can you show us any example or pictorial example?

[Aryana]

Well I have an element represented by indices α, β, γ in a K*K*K array and I need to find out corresponding points in my 3D meshgrid. But I may be thinking about it incorrectly and it might rely on some Fourier Transform intuition that I don't have right now.

Answer 1

[John D'Errico]

You already have the coordinates, so I'm not at all sure what your question asks. Perhaps an example would help.

x = -10:5:10

x = 1×5

-10 -5 0 5 10

y = 0:5:25

y = 1×6

0 5 10 15 20 25

z = -20:10:20

z = 1×5

-20 -10 0 10 20

Now use meshgrid.

[X,Y,Z] = meshgrid(x,y,z);

This creates 3 arrays, each of size

size(X)

ans = 1×3

6 5 5

So y varies the fastest. (Personally, this has always bugged me, so I have always preferred ndgrid. The difference is not really that important though, and there was a valid reason for that choice.) X looks like this:

X

X = 
X(:,:,1) =

   -10    -5     0     5    10
   -10    -5     0     5    10
   -10    -5     0     5    10
   -10    -5     0     5    10
   -10    -5     0     5    10
   -10    -5     0     5    10


X(:,:,2) =

   -10    -5     0     5    10
   -10    -5     0     5    10
   -10    -5     0     5    10
   -10    -5     0     5    10
   -10    -5     0     5    10
   -10    -5     0     5    10


X(:,:,3) =

   -10    -5     0     5    10
   -10    -5     0     5    10
   -10    -5     0     5    10
   -10    -5     0     5    10
   -10    -5     0     5    10
   -10    -5     0     5    10


X(:,:,4) =

   -10    -5     0     5    10
   -10    -5     0     5    10
   -10    -5     0     5    10
   -10    -5     0     5    10
   -10    -5     0     5    10
   -10    -5     0     5    10


X(:,:,5) =

   -10    -5     0     5    10
   -10    -5     0     5    10
   -10    -5     0     5    10
   -10    -5     0     5    10
   -10    -5     0     5    10
   -10    -5     0     5    10

So each array has 3 dimensions. But we can also convert them to a list of (x,y,z) triads, as:

XYZ = [X(:),Y(:),Z(:)]

XYZ = 150×3

-10 0 -20 -10 5 -20 -10 10 -20 -10 15 -20 -10 20 -20 -10 25 -20 -5 0 -20 -5 5 -20 -5 10 -20 -5 15 -20

Each row of the resulting array corresponds to one (x,y,z) triple. There will be 5x6x5=150 of those triples.

size(XYZ)

ans = 1×2

150 3

Anyway, it will help once you start learning to use and write vectorized codes, where X, Y, and Z are all arrays. In that sense (X,Y,Z) really are the vector coordinates you are looking for.

[Aryana]

This was very helpful thank you. I realized that what I needed to do was already accomplished when I used ind2sub to get the indices, so you are right that I already had what I needed.

Answer 2

[Benjamin Campbell]

Presuming you start with vectors x, y and z. When you do [X Y Z] = meshgrid(x, y, z) you get three 3D arrays with every combination. To answer the question it depends on how you get the indices (x1 y1 z1). If you get them from the original vectors, you can just do x(x1) y(y1) z(z1). Or, if you get them from the meshgrid values it will be X(x1, y1, z1) Y(x1, y1, z1) Z(x1, y1, z1)

If you share code it will be easier to answer, but hope that helps.

[Aryana]

Thank you this was instructive!