Can querying "TriScatteredInterp" be as fast as "griddedInterpolant"?

Question

dominik 2013-7-9

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/81578-can-querying-triscatteredinterp-be-as-fast-as-griddedinterpolant

Hi, im wondering why querying the TriScatteredInterpolant is so much slower and so much more size dependent than griddedInterpolant. Is there a way to have the fast performance of the griddedinterpolant for scattered data interpolation?

For illustrative purpose consider this example, where i create a gridded and a scattered interpolant for a smaller grid and a larger grid

[x1 x2 x3]=ndgrid(1:30,1:30,1:30);
[y1 y2 y3]=ndgrid(1:100,1:100,1:100);
x4=(x1+x2+x3);
y4=(y1+y2+y3);
LLip1  = griddedInterpolant(x1,x2,x3,x4)  ;
LLip2  = griddedInterpolant(y1,y2,y3,y4)  ;
LLip4  = TriScatteredInterp(y1(:),y2(:),y3(:),y4(:))  ;
LLip3  = TriScatteredInterp(x1(:),x2(:),x3(:),x4(:))  ;
display('small gridded')
tic; for i=1:100; LLip(2.5,3.4,3.2); end; toc
display('large gridded')
tic; for i=1:100; LLip2(2.5,3.4,3.2); end; toc
display('small scattered')
tic; for i=1:100; LLip3(2.5,3.4,3.2); end; toc
display('large scattered')
tic; for i=1:100; LLip4(2.5,3.4,3.2); end; toc

Output (matlab 2012a64bit): small gridded Elapsed time is 0.002512 seconds. large gridded Elapsed time is 0.002306 seconds. small scattered Elapsed time is 0.281428 seconds. large scattered Elapsed time is 9.635930 seconds.

(In my application in fact, similaly to the example, my data is not properly scattered. It merely has ragged edges, like

 2   1   NaN
 1   NaN NaN
 NaN NaN NaN

i thought about using the griddedinterpolant when possible and using the sctatteredInerpolant only for the edges, i.e. when the gridded would return NaN, but a fast triscatteredinterp would make things easier...)

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Matt J 2013-7-9

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/81578-can-querying-triscatteredinterp-be-as-fast-as-griddedinterpolant#answer_91282

编辑：Matt J 2013-7-9

Is there a way to have the fast performance of the griddedinterpolant for scattered data interpolation?

I doubt it. gridded interpolation is fundamentally easier because data neighbors that participate in interpolation at a given point are easy to find due to their plaid arrangement. Additionally, the interpolation operations are tensorially separable.

Conversely, with scattered interpolation, it's harder to figure out which neighbours should participate. E.g., TriScatteredInterp uses a Delaunay triangulation to determine this. Additionally, the interpolation is not tensorially separable in the scattered case.

7 个评论
显示 5更早的评论隐藏 5更早的评论

dominik 2013-7-12

Hey Matt, thanks for making the effort to try to understand my problem!

Well, unfortunatley I dont have a closed from solution, but i know the domain where the function is defined. And I can evaluate the function at any point using a complicate algorithm (basically solving a huge system of nonlinear equations). But this is very (prohibitively) time consuming (so i want to get the values of the function on some grid and then use an interpolant to save time).

By taking samples closer and closer to the boarder of the domain i observed the function seems to go to infinity (and this matches the intuition of what the function should do).