How to 3d parameter spline

[wssyj]

Sorry to disturb you.I am a beginner for your library, after seeing the part of parameter spline, I would like to ask how to use 3d parameter spline.

Here is your two-dimensional example：

using Interpolations
t = 0:.1:1
x = sin.(2π*t)
y = cos.(2π*t)
A = hcat(x,y)
itp = Interpolations.scale(interpolate(A, (BSpline(Cubic(Natural(OnGrid()))), NoInterp())), t, 1:2)
tfne = 0:.01:1
xs, ys = [itp(t,1) for t in tfine], [itp(t,2) for t in tfine]

[mkitti]

julia> A = rand(4,4,4)
4×4×4 Array{Float64, 3}:
[:, :, 1] =
 0.602447  0.961579  0.302673  0.0647496
 0.93965   0.959496  0.227365  0.66319
 0.331185  0.100806  0.920701  0.215262
 0.608488  0.518511  0.896318  0.668993

[:, :, 2] =
 0.081988  0.970282   0.306302  0.868053
 0.959003  0.414357   0.491382  0.385106
 0.247292  0.610567   0.629706  0.779416
 0.467251  0.0550997  0.405658  0.995051

[:, :, 3] =
 0.310974  0.840416  0.492628   0.490059
 0.510787  0.110757  0.0374218  0.643901
 0.509094  0.551814  0.289287   0.675418
 0.81481   0.845827  0.275486   0.314155

[:, :, 4] =
 0.67629   0.683418  0.645005  0.285094
 0.650935  0.550022  0.237134  0.152433
 0.368105  0.689606  0.708606  0.647533
 0.669374  0.76532   0.698094  0.789102

julia> nodes = ([x^2 for x in 1:4], [0.2y for y in 1:4], [z^3/2 for z in 1:4])
([1, 4, 9, 16], [0.2, 0.4, 0.6000000000000001, 0.8], [0.5, 4.0, 13.5, 32.0])

julia> itp = interpolate(nodes, A, Gridded(Linear()))
4×4×4 interpolate((::Vector{Int64},::Vector{Float64},::Vector{Float64}), ::Array{Float64, 3}, Gridded(Linear())) with element type Float64:
[:, :, 1] =
 0.602447  0.961579  0.302673  0.0647496
 0.93965   0.959496  0.227365  0.66319
 0.331185  0.100806  0.920701  0.215262
 0.608488  0.518511  0.896318  0.668993

[:, :, 2] =
 0.081988  0.970282   0.306302  0.868053
 0.959003  0.414357   0.491382  0.385106
 0.247292  0.610567   0.629706  0.779416
 0.467251  0.0550997  0.405658  0.995051

[:, :, 3] =
 0.310974  0.840416  0.492628   0.490059
 0.510787  0.110757  0.0374218  0.643901
 0.509094  0.551814  0.289287   0.675418
 0.81481   0.845827  0.275486   0.314155

[:, :, 4] =
 0.67629   0.683418  0.645005  0.285094
 0.650935  0.550022  0.237134  0.152433
 0.368105  0.689606  0.708606  0.647533
 0.669374  0.76532   0.698094  0.789102

julia> itp(2, 0.5, 2)
0.600909185850235

See http://juliamath.github.io/Interpolations.jl/latest/control/#Gridded-interpolation

Does this help?

[wssyj]

Thank you very much！！！

[wssyj]

Sorry to bother you again, but I have one more question For this example

using Interpolations
t = 0:.1:1
x = sin.(2π*t)
y = cos.(2π*t)
A = hcat(x,y)
itp = Interpolations.scale(interpolate(A, (BSpline(Cubic(Natural(OnGrid()))), NoInterp())), t, 1:2)
tfne = 0:.01:1
xs, ys = [itp(t,1) for t in tfine], [itp(t,2) for t in tfine]

x is not unique and sorted in increasing order.

How do I interpolate if the nodes in your example are not unique and not additive For example

nodes = ([1,3,1,0], [1,3,1,0], [z^3/2 for z in 1:4])

[mkitti]

If they are not sorted, then you can sort them. If the nodes are not unique, I'm not clear what you expect the value to be at the nodes.