ScatteredInterpolant for Surface Extrapolation

Question

MATLAB_Soldier 2022-9-21

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1809200-scatteredinterpolant-for-surface-extrapolation

移动：Bruno Luong 2022-9-21

Hi Everyone,

I have a set of XYZ data that forms the following plot (surface, not in this case) below.I would like to extrapolate this slightly outside of the region on the X axis, at 2000.

I have used the following code to extrapolate at the value of 2000.

%Input for Extrapolation
xq=zeros(21,1)+2000;
yq=[0:5:95]';
yq(end+1)=99;
F = scatteredInterpolant(MAP_X,MAP_Y,MAP_Z);
vq1=F(xq,yq)

Unfortuntaly, my results are far off from ideal:

vq1 =

0.4740

305.5350

130.4689

-156.4641

-19.2672

86.5564

281.8144

112.3261

-150.5571

534.2845

-219.2750

245.6169

-219.3640

60.4432

-141.4669

252.9336

-181.9238

115.2369

-131.9798

58.5419

4.9041

What is causing this error? How can I fix it?

Is there a better way to extrapolate this dataset?

Many thansk.

5 个评论
显示 3更早的评论隐藏 3更早的评论

KSSV 2022-9-21

编辑：KSSV 2022-9-21

Extrapolation cannot be trusted. You specify the method as nearest.

MATLAB_Soldier 2022-9-21

Nearest method gives poor results. It is very unrealistic. Look at the plot below, values extrapolated to 2000.

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Matt J 2022-9-21

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1809200-scatteredinterpolant-for-surface-extrapolation#answer_1057765

编辑：Matt J 2022-9-21

在 MATLAB Online 中打开

The surface appears to be very well modeled by a Gaussian. An appropriate thing would therefore be to extrapolate using a Gaussian fit, e.g., using this FEX download,

https://www.mathworks.com/matlabcentral/fileexchange/69116-gaussfitn

[MAP_X,MAP_Y,MAP_Z]=readvars('https://www.mathworks.com/matlabcentral/answers/uploaded_files/1131415/MATLAB.csv');

res=@(q)reshape(q,21,[])';

MAP_X=res(MAP_X);

MAP_Y=res(MAP_Y);

MAP_Z=res(MAP_Z);

params=gaussfitn([MAP_X(:),MAP_Y(:)],MAP_Z(:));

Local minimum possible. lsqcurvefit stopped because the size of the current step is less than the value of the step size tolerance.

[D,A,mu,sigma]=deal(params{:});

fcn=@(x,y) D+A*exp(-0.5*sum( ([x-mu(1),y-mu(2)]/sigma).*[x-mu(1),y-mu(2)] ,2));

xq=zeros(21,1)+2000;

yq=[0:5:95]';

yq(end+1)=99;

vq1=fcn(xq,yq)

vq1 = 21×1

0.2207 0.6207 1.0277 1.4359 1.8391 2.2305 2.6032 2.9503 3.2649 3.5407

scatter3(MAP_X(:),MAP_Y(:),MAP_Z(:)); hold on

scatter3(xq,yq,vq1); hold off

xlabel x

ylabel y

view(-60,15)

1 个评论
显示 -1更早的评论隐藏 -1更早的评论

MATLAB_Soldier 2022-9-21

移动：Bruno Luong 2022-9-21

Thank you very much for your help guys and for the excellent explanation Walter Robinson.

请先登录，再进行评论。

Answer 2

Walter Roberson 2022-9-21

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1809200-scatteredinterpolant-for-surface-extrapolation#answer_1057755

scatteredInterpolant() does not do any kind of surface fitting. The default fitting option that it uses, the one you are getting, is 'linear'. A Delaunay triangulation is done, nearest points on the triangulation found, linear interpolation is done.

Depending on the input coordiantes and the query coordinates, it is not uncommon for the nearest point in the triangulation to be different than expected. It is common for large triangles in Z to be created that cross a lot of the surface. The range of Z values in your surface is small compared to the range of X values.

Near the edge of the surface, extrapolating outside... the interior behaviour of the surface can become pretty much irrelevant. If you do not end up with one of those large spanning triangles just mentioned, if you end up querying against a more local series of triangles, then the rest of the surface might as well not exist. And that implies that small differences in Z values near the outside edge can get magnified to large differences when you query a distance away. For example if you are querying against a couple of surface points that differ by 1/2 in Z coordinates, and you are querying 1000 units away, then by linear interpolation the predicted Z difference would be 500, even though that is much larger than any of the input Z values.

If you could rewrite in terms of griddedInterpolant (since you pretty much have a grid of X and Y) then you would be able to use 'spline' or 'pchip' interpolant.