3D look up interp3

May I know, I have a problem with 3D lookup table. I have 16x40x3 table, all populated. But the values ​​I want to have 16x79x9 values ​​which all exists in the limits. And I am using interp3 (x, y, z, V, x ', y', z '). And the sizes are

x: 16 valued array

y: 40 valued array

z: 3 valued array

V: 16x40x3 matrix

x ', y', z 'are scalars

May I know what is wrong here? I find the error.

Error using gridded Interpolant

The grid vectors do not define a grid of points that matches the given values.

Please help

Answer 1

[David Wilson]

You don't give us much to work on, so I'll make up some data that follow your dimensions.

x = linspace(-1,1,16); y =linspace(-1,1,40); z = linspace(-1,1,3); 
[X,Y,Z] = meshgrid(x,y,z);
V = sin(X)+cos(Y)+Z.^2; % some function 

This generates a data cube in the dimensions you said (16*40*3). I've made up a well behaved function (since I had nothing to go on.)

Now we will generate an interpolating grid cube of the dimensions you wanted (16*79*9) using meshgrid.

xi = linspace(-0.5,1,16); yi = linspace(-0.8,0.8,79); zi = linspace(-1,1,9); 
[Xi,Yi,Zi] = meshgrid(xi,yi,zi); 

Now we are ready to interpolate in 3D.

Vi = interp3(X,Y,Z,V, ... 
             Xi, Yi, Zi)

The variable Vi is your interpolated output.

>> size(Vi)
ans =
    79    16     9

[Sivateja Maturu]

In my case I am looking for a interpolation of scalars. I mean,xi, yi,zi are scalrs in my case which are within limits of x,y,z

Answer 2

[David Wilson]

In that case it's easy, just use them as the interpolated input: say zt x=0.5, y=0.7,z=-0.3

Vi = interp3(X,Y,Z,V, ... 
             0.5, 0.7, -0.3)

[Gabriel Felix]

interp3(X,Y,Z,V,Xq,Yq,Zq):

- X is the variable that contains the base values of the columns of V.

- Y is the variable that contains the base values of the rows of V.

- Z is the variable that contains the base values of the 3rd dimension of V

The names of variables X, Y and Z further confuses us, because they dont mean the literal vectors X, Y and Z. They express the reference values for:

X: Columns - disposed on the rows

Y: Rows - disposed on the columns

Z: 3rd dimension - disposed on the 3rd dimension

Ex.

V = [1 2 3 4 5 ;

6 7 8 9 10]

sizeX = 5 -> 5 columns

sizeY = 2 -> 2 rows

sizeZ = 1 -> 1 3rd dim

interp3(V,5,2,1) = 10

Ex.

V(:,:,1) = [

1 2 3 4

5 6 7 8

9 10 11 12]

V(:,:,1) = [

13 14 15 16

17 18 19 20

21 22 23 24]

sizeX = 4 -> 4 columns

sizeY = 3 -> 3 rows

sizeZ = 2 -> 2 3rd dim

interp3(V,3,3,2) = 19

Obs.: I didnt use the X,Y,Z values because I would have to create them and it would take long

The same applies to interp2