Better method to interpn when one grid is fixed?

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Sean
Sean 2017-3-3
评论： Sean 2017-3-7
I currently have a n-D sized database, and I'm using interpn to interpolate values from it. However, I do not interpolate on the last dimension, as I'm using the database to interpolate between 1xn solutions. The database is very large, and I am interpolating on it many times, so I'm looking for a faster solution. I believe there should be a faster solution since I'm not necessarily "interpolating" between values on the last dimension. Below is an example of my current method that I am looking to improve:
Example:
database=rand([4,4,2,3,8,6,2,6,5500]);
interpGrids={[1000 2000 4000 7000],[-5 -2 2 5],[0 1],[-5 0 5],[-20 -15 -5 0 5 10 15 20],[10 20 30 40 50 60],[-5 0],[.1 .2 .3 .4 .5 1],1:5500};
ngGriddedData=cell(1,9);
[ngGriddedData{:}]=ndgrid(interpGrids{:});
tic
solution=interpn(ngGriddedData{:},database,3000,1,1,2,-17,35,-4,.35,1:5500);
toc
Many many thanks for any suggestions. I'll be sure to send some good computing karma your way.
- Sean

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采纳的回答

Prashant Arora
Prashant Arora 2017-3-6
Hi Sean,
You can use the griddedInterpolant function, which provides performance improvements for repeated queries to the interpolant.
F = griddedInterpolant(interpGrids,database);
tic
solution2 = F({3000,1,1,2,-17,35,-4,.35,1:5500});
toc

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