# Interpolate a 3D-Matrix using interp1

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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!
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### Réponse acceptée

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.
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### Plus de réponses (1)

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
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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 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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