How to extrapolate on non-uniform grid?

Specific Domains Numerics
1

How to interpolate(actually extrapolate later) Bp values on r,θand ϕ grid? I have non-uniform grid of r,θand ϕ, where r is geometric series and θand ϕ have constant spacing.

r = Float32[1.677044, 1.8385518, 2.0156136, 2.2097273, 2.4225352, 2.6558375, 2.911608, 3.1920104, 3.499417, 3.8364286, 4.205896, 4.6109447, 5.055002, 5.5418243, 6.0755296, 6.660634, 7.3020864, 8.005314];
θ = Float32[0.049087387, 0.14726216, 0.24543692, 0.3436117] ;
ϕ = Float32[0.19634955, 0.5890486, 0.9817477, 1.3744467, 1.7671459, 2.1598449, 2.552544, 2.9452431, 3.3379421, 3.7306414, 4.12334, 4.5160394, 4.9087386, 5.3014374, 5.6941366, 6.086836];
#= This code prints geometric ratio of r
for i in 2:length(r)-1
   println(r[i]/r[i-1])
end=#
Bp = rand(18, 4, 16)
2

If r is geometric, then log r should have constant spacing, so you can do interpolation/extrapolation on a uniform grid using this change of variables.

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3 
ext_itp = extrapolate(interpolate((log.(r), θ, ϕ), Bp, Gridded(Linear())), Interpolations.Line());
ext_itp(10, 2,3)
ext_itp = extrapolate(interpolate((r, θ, ϕ), Bp, Gridded(Linear())), Interpolations.Line());
ext_itp(10, 2,3)

Yes, log.(r) will make it uniform, but it will be extrapolation on different value. Note that i am using Polar coordinates. How should i call ext_itp() on values?

4

As @JADekker suggested, you can define your extrapolator function (functor) like this:

using Interpolations
extrapol = linear_interpolation((log.(r), θ, ϕ), Bp, extrapolation_bc=Line())

And then evaluate it at any point r₁, θ₁, ϕ₁ as follows:

extrapol(log(r₁), θ₁, ϕ₁)
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