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Interpolate a 3D-Matrix using interp1

조회 수: 17 (최근 30일)
Matthias
Matthias 2012년 12월 8일
편집: Matt J 2014년 1월 16일
I try to interpolate a 3D Matrix using interp1. I think it should be easy but i just don't get it.
I have a 3d-matrix and a vector in z direction. I wan't to get an interpolated 2d Matrix for any value in the z direction. For Example:
matrix(1:3,1:3,1) = 1
matrix(1:3,1:3,2) = 2
z(1,1,1:2) = [10,20]
interp1(z,matrix,15)
so I thought interp1 would return a 3x3 Matrix with the values 1.5 but i just get an error.
Thanks a lot!

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Matt J
Matt J 2012년 12월 8일
편집: Matt J 2012년 12월 8일
V=reshape(matrix,[],2).';
VI = interp1(z,V,15);
ans = reshape(VI,3,3),
You could also use griddedInterpolant instead
F=griddedInterpolant({1:3,1:3,z},matrix);
ans= F({1:3,1:3,15});
It's hard to say which would be more efficient, in general. griddedInterpolant is a newer and better optimized function, but interp1 takes advantage of the 1D nature of the interpolation.

추가 답변 (1개)

Glynn
Glynn 2014년 1월 15일
Using the same example above, what if the xi in interp1(x,Y,xi), is a 2-d matrix the same size as the first two dimensions of 'matrix'? I keep getting OOM issues when trying to solve: x is a (1,3)array Y is a (2030,873,3) matrix xi is a (2030,873) matrix
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Glynn
Glynn 2014년 1월 16일
Okay got it, thanks, makes sense. So should I be using interp3 instead? I'm confused as how to set that up with my particular example. Any help is appreciated.
Matt J
Matt J 2014년 1월 16일
편집: Matt J 2014년 1월 16일
Yes, you could use interp3, though I think griddedInterpolant is easier to work with. I don't actually have a clear picture how your xi are supposed to be interpreted. If you are interpolating in a 3D array, like Y, then the locations at which you interpolate must consist of (xx,yy,zz) coordinate triplets.

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