正确使用scipy.interpolate.RegularGridInterp

Question

我有点被documentation for scipy.interpolate.RegularGridInterpolator搞糊涂了。

例如，我有一个函数f:R^3=>；R，它在单位立方体的顶点上采样。我想插值以便在多维数据集中找到值。

import numpy as np

# Grid points / sample locations
X = np.array([[0,0,0], [0,0,1], [0,1,0], [0,1,1], [1,0,0], [1,0,1], [1,1,0], [1,1,1.]])

# Function values at the grid points
F = np.random.rand(8)

现在，RegularGridInterpolator接受一个points参数和一个values参数。

points : tuple of ndarray of float, with shapes (m1, ), ..., (mn, ) The points defining the regular grid in n dimensions.

values : array_like, shape (m1, ..., mn, ...) The data on the regular grid in n dimensions.

我将此解释为能够这样调用：

import scipy.interpolate as irp

rgi = irp.RegularGridInterpolator(X, F)

但是，当我这样做时，会出现以下错误：

ValueError: There are 8 point arrays, but values has 1 dimensions

我在文件里误解了什么？

Answer 1

你的回答更好，你完全可以接受。我只是添加这个作为一个“替代”的方式来编写它。

import numpy as np
import scipy.interpolate as spint

RGI = spint.RegularGridInterpolator

x = np.linspace(0, 1, 3) #  or  0.5*np.arange(3.) works too

# populate the 3D array of values (re-using x because lazy)
X, Y, Z = np.meshgrid(x, x, x, indexing='ij')
vals = np.sin(X) + np.cos(Y) + np.tan(Z)

# make the interpolator, (list of 1D axes, values at all points)
rgi = RGI(points=[x, x, x], values=vals)  # can also be [x]*3 or (x,)*3

tst = (0.47, 0.49, 0.53)

print rgi(tst)
print np.sin(tst[0]) + np.cos(tst[1]) + np.tan(tst[2])

返回：

1.93765972087
1.92113615659