4d Interpolation of Table

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River Vorse
River Vorse 2020 年 11 月 9 日
回答済み: Raunak Gupta 2020 年 11 月 13 日
I'm trying to interpolate a set of 4 variable data values in a table. Can I use interpn() for this?
the first column ranges (0:2:10)- full table not shown since there are hundreds of values. The first three columns are inputs such that at x =0, y= 16, z=40, there are a number of outputs shown in the right 7 columns. I'm trying to interpolate so that if I have a set of inputs x = 0.5, y = 16.7, z = 37.2, I can interpolate a value for one of the outputs in the right 7 columns.
  2 件のコメント
River Vorse
River Vorse 2020 年 11 月 9 日
My thought is :
x = data(:,1);
y = data(:,2);
z = data(:,3);
v = data(;,4);
xp = 0.5;
yp = 16.7;
zp = 37.2;
vp = interpn(x,y,z,v,xp,yp,zp)
River Vorse
River Vorse 2020 年 11 月 9 日
this gives the following error

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回答 (1 件)

Raunak Gupta
Raunak Gupta 2020 年 11 月 13 日
Hi,
The error when using interpn is due to the data not being present in Grid format, meaning for every triplet of (x,y,z) in some range of x,y,z should have a value v also present. You can create the Grid format data for x,y,z using
[X1, X2, X3] = ndgrid(x,y,z);
But for the same you need to reshape the 4th column such that it represents corresponding values in X1, X2, X3. For this the length of unique x,y,z is required which gives the indices where values for learning interpolation are present.
You can see here about how the input array should look like.

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