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Interpolate a 3D-Matrix using interp1

17 vues (au cours des 30 derniers jours)
Matthias
Matthias le 8 Déc 2012
Modifié(e) : Matt J le 16 Jan 2014
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!

Réponse acceptée

Matt J
Matt J le 8 Déc 2012
Modifié(e) : Matt J le 8 Déc 2012
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.

Plus de réponses (1)

Glynn
Glynn le 15 Jan 2014
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
  3 commentaires
Glynn
Glynn le 16 Jan 2014
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 le 16 Jan 2014
Modifié(e) : Matt J le 16 Jan 2014
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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