scipy.interpolate.RegularGridInterpolator 的正确用法

Question

我对 documentation for scipy.interpolate.RegularGridInterpolator 有点困惑。

比如说我有一个函数 f: R^3 => R 是在单位立方体的顶点上采样的。我想插值以便在立方体中找到值。

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 个参数，一个 values 个参数。

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.

我将其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，出现以下错误：

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

我在文档中误解了什么？

Answer 1

好吧，当我回答自己的问题时，我觉得自己很傻，但我在原始 regulargrid 库的文档的帮助下发现了我的错误：

https://github.com/JohannesBuchner/regulargrid

points 应该是一个数组列表，指定点沿每个轴的间隔方式。

比如上面的单位立方体，我应该设置：

pts = ( np.array([0,1.]), )*3

或者如果我有沿最后一个轴以更高分辨率采样的数据，我可能会设置：

pts = ( np.array([0,1.]), np.array([0,1.]), np.array([0,0.5,1.]) )

最后，values 的形状必须与 points 隐式布局的网格相对应。例如，

val_size = map(lambda q: q.shape[0], pts)
vals = np.zeros( val_size )

# make an arbitrary function to test:
func = lambda pt: (pt**2).sum()

# collect func's values at grid pts
for i in range(pts[0].shape[0]):
    for j in range(pts[1].shape[0]):
        for k in range(pts[2].shape[0]):
            vals[i,j,k] = func(np.array([pts[0][i], pts[1][j], pts[2][k]]))

最后，

rgi = irp.RegularGridInterpolator(points=pts, values=vals)

根据需要运行和执行。

Answer 2

你的回答比较好，完全可以采纳。我只是将其添加为 "alternate" 编写脚本的方式。

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])

returns:

1.93765972087
1.92113615659

scipy.interpolate.RegularGridInterpolator 的正确用法

Correct usage of scipy.interpolate.RegularGridInterpolator

python

interpolation

numpy

scipy