Spline interpolation in 3D in python

Question

I am searching the equivalent Matlab command

Vq = interp3(X,Y,Z,V,Xq,Yq,Zq)

in Python. In Matlab I can use the method 'spline' interpolation, which I can not find in python for 3D data. There exists scipy.interpolate.griddata, but it doesn't have the option spline for 3D data.

The data I want to interpolate is a 3D matrix (51x51x51), which is regularly distributed on a 3D grid.

scipy.interpolate.Rbf may be the option, but I don't get it working:

xi = yi = zi = np.linspace(1, 132651, 132651) interp = scipy.interpolate.Rbf(xi, yi, zi, data, function='cubic')

leads to a memory error.

Edit: A minimal example of what I want (without interpolation): Matlab code

v=rand([51,51,51]);
isosurface (v, 0.3);

For simplicity, I use random data in this example. I want to make isosurface plots (in particular, Fermi surface plots). Since some structures are very small, a high grid resolution of 51x51x51 is needed.

A further comment: The data set in the matrix is independent from each other, z (or the 3rd component) is NOT a function of x and y.

Answer 1

The mess example worked great. I want to create a b spline of the data in 3d

data = np.array([[ 41. ,  57. ,  92. ],[ 39. ,  57.  , 92.4],[ 43. ,  57.  , 91.2], [ 23.,   47. , 119.6],
                 [ 27. ,  47. , 115.2], [ 25. ,  45. , 122. ], [ 25. ,  49. , 114. ],[ 29.,   49. , 109.6],
                 [ 29. ,  47. , 114.4], [ 27. ,  49. , 111.2], [ 23. ,  45. , 125.6], [ 31.,   49.,  106.8],
                 [ 25. ,  47. , 117.6], [ 39. ,  55. ,  95.6],[ 37.  , 53.  , 98.4], [ 35. ,  55. ,  96.8],
                 [ 33. ,  53. , 116.8], [ 23. ,  43. , 132.8], [ 25. ,  41. , 145.2],[ 25. ,  43.,  133.6],
                 [ 29. ,  51. , 106.4],[ 31.  , 53. , 121.2],[ 31., 51. , 104.8],[ 41.,   55.,   93.6],
                 [ 33. ,  51. , 103.6],[ 35.  , 53. ,  99.6],[ 37. ,  55. ,  96.4]])

x = data[:,0]
y = data[:,1]
z = data[:,2]

# sort data to avoid plotting problems
x, y, z = zip(*sorted(zip(x, y, z)))

x = np.array(x)
y = np.array(y)
z = np.array(z)

import scipy as sp
import scipy.interpolate
from mpl_toolkits.mplot3d import Axes3D

spline = sp.interpolate.Rbf(x,y,z,function='thin_plate',smooth=5, episilon=5)

x_grid = np.linspace(min(x),max(x), len(x))
y_grid = np.linspace(min(y),max(y), len(y))
B1, B2 = np.meshgrid(x_grid, y_grid, indexing='xy')

Z = spline(B1,B2)
fig = plt.figure(figsize=(10,6))
ax = Axes3D(fig)
ax.plot_wireframe(B1, B2, Z)
ax.plot_surface(B1, B2, Z,alpha=0.2)
ax.scatter3D(x,y,z, c='r')
plt.show()

Answer 2

Spline interpolation on for 3+ dimensions can be done using scipy.interpolate.Rbf as your described. For plotting purposes you can use a smaller resolution (1000 points is a good rule of thumb), and when you want to evaluate your spline, you can interpolate on much greater than 132000 points without problem (see example below).

Can you add a Minimal, Complete, and Verifiable example for what you are trying to do in matlab? This will explain why do you need to create a grid space with a resolution of 132000 points. Also, please note, there is a curse of dimensionality. Matlab uses a cubic spline or a piecewise polynomial which can be dangerous due to overfitting. I recommend you used a more sane method for training on 51 datapoints and applying to 132000+ datapoints. This is a great example on polynomial curve fitting and model selection.

Example:

Generate data:

import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import axes3d

%matplotlib inline
import random
# set seed to reproducible
random.seed(1)
data_size = 51
max_value_range = 132651
x = np.array([random.random()*max_value_range for p in range(0,data_size)])
y = np.array([random.random()*max_value_range for p in range(0,data_size)])
z = 2*x*x*x + np.sqrt(y)*y + random.random()
fig = plt.figure(figsize=(10,6))
ax = axes3d.Axes3D(fig)
ax.scatter3D(x,y,z, c='r')

Fit spline and interpolate

x_grid = np.linspace(0, 132651, 1000*len(x))
y_grid = np.linspace(0, 132651, 1000*len(y))
B1, B2 = np.meshgrid(x_grid, y_grid, indexing='xy')
Z = np.zeros((x.size, z.size))

import scipy as sp
import scipy.interpolate
spline = sp.interpolate.Rbf(x,y,z,function='thin_plate',smooth=5, episilon=5)

Z = spline(B1,B2)
fig = plt.figure(figsize=(10,6))
ax = axes3d.Axes3D(fig)
ax.plot_wireframe(B1, B2, Z)
ax.plot_surface(B1, B2, Z,alpha=0.2)
ax.scatter3D(x,y,z, c='r')

Fit spline on large data

predict_data_size = 132000
x_predict = np.array([random.random()*max_value_range for p in range(0,predict_data_size)])
y_predict = np.array([random.random()*max_value_range for p in range(0,predict_data_size)])
z_predict = spline(x_predict, y_predict)
fig = plt.figure(figsize=(10,6))
ax = axes3d.Axes3D(fig)
ax.plot_wireframe(B1, B2, Z)
ax.plot_surface(B1, B2, Z,alpha=0.2)
ax.scatter3D(x_predict,y_predict,z_predict, c='r')