3D look up interp3

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Sivateja Maturu
Sivateja Maturu 2019-4-9
编辑： Gabriel Felix 2021-4-28
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

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回答（2 个）

David Wilson
David Wilson 2019-4-9
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
  1 个评论
Sivateja Maturu
Sivateja Maturu 2019-4-9
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

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David Wilson
David Wilson 2019-4-9
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)
  1 个评论
Gabriel Felix
Gabriel Felix 2021-4-28
编辑：Gabriel Felix 2021-4-28
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

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