How can one create this matrix, part 2

In the previous post, I have learned that

 [X, Y] = meshgrid(0:.1:0.5);
 [X(:), Y(:)]
 

will create a matrix

 0 0 
 0 0.1
 0 0.2
 ...
     

Now I want to generalize to

 [X, Y, Z] = meshgrid(0:.1:0.5);
 [X(:), Y(:), Z(:)]

and it creates a matrix

ans = 216×3
         0         0         0
         0    0.1000         0
         0    0.2000         0
         0    0.3000         0
         0    0.4000         0
         0    0.5000         0
    0.1000         0         0
    0.1000    0.1000         0
    0.1000    0.2000         0
    0.1000    0.3000         0

But the matrix I want to create is like

0 0 0 
0 0 0.1
0 0 0.2
0 0 0.3
0 0 0.4
0 0 0.5
0 0.1 0.1
0 0.1 0.2
0 0.1 0.3

Please advise.

Answer 1

[madhan ravi]

 [X, Y, Z] = ndgrid(0:.1:0.5);
 [Z(:), Y(:), X(:)]

[alpedhuez]

Thank you. But how does it work?

[madhan ravi]

 doc ndgrid

[Bruno Luong]

Thank you. But how does it work?

Just think those combinations you look for

(1,1)

(1,2)

...

(1,5)

(2,1)

...

(5,5)

as 2D coordinates of a set of grid points in a rectangles (1:5) x (1:5).

For 3 combinations, it's a 3D coordinates of grid points of a cube (1:5) x (1:5) x (1:5),

ndgrid() function generates grid points in n-dimensional space.

[alpedhuez]

Yes. But whey (Z,Y,X) not (X,Y,Z)?

[Bruno Luong]

Because NDGRID returns result such that the first input change first (more rapidly) and you want the opposite, X change most slowly. Actually the code as written by madhan is confusing, it's clearer this way to me:

 [Z,Y,X] = ndgrid(0:.1:0.5);
 [X(:), Y(:), Z(:)]

You'll see if 1D-grid of X, Y, Z are different and not all equal to 0.0.1:0:5, it makes a whole world less confusing.