正确使用scipy.interpolate.RegularGridInterpolator [英] Correct usage of scipy.interpolate.RegularGridInterpolator

问题描述

有关scipy的文档让我有些困惑.interpolate.RegularGridInterpolator .

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.我想进行插值以便在多维数据集中找到值.

Say for instance I have a function f: R^3 => R which is sampled on the vertices of the unit cube. I would like to interpolate so as to find values inside the cube.

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使用一个points参数和一个values参数.

Now, RegularGridInterpolator takes a points argument, and a values argument.

点:浮点数ndarray的元组，形状为(m1，)，...，(mn，) 在n个维度上定义规则网格的点.

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

值:类数组，形状(m1，...，mn，...) 规则网格中n个维度上的数据.

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

我认为这可以打电话:

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:

However, when I do so, I get the following error:

ValueError:有8个点数组，但值具有1个维度

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

我在文档中误解了什么?

What am I misinterpreting in the docs?

推荐答案

好的，当我回答自己的问题时，我感到很傻，但是在原始regulargrid lib的文档的帮助下，我发现了自己的错误:

Ok I feel silly when I answer my own question, but I found my mistake with help from the documentation of the original regulargrid lib:

https://github.com/JohannesBuchner/regulargrid

points应该是一个数组列表，用于指定沿每个轴的点间距.

points should be a list of arrays that specifies how the points are spaced along each axis.

例如，要采用上述单位立方体，我应该设置:

For example, to take the unit cube as above, I should set:

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

或者如果我有沿最后一个轴以更高的分辨率采样的数据，则可以设置:

or if I had data which was sampled at higher resolution along the last axis, I might set:

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，

Finally, values has to be of shape corresponding to the grid laid out implicitly by points. For example,

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)

根据需要运行并执行.

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