如何在网格中对点列表进行二维插值 [英] How to do 2d interpolation in a grid for a list of points

问题描述

似乎这里的一些相关问题没有给我想要的答案.

It seems a few relevant questions here didn't give me the answer I want.

所以我有一个 2D 查找表，它是一个 2D 网格，其值位于网格坐标处.例如，可以通过以下代码生成这样的表:

So I have a 2D look up table which is a 2D grid with values at grid coordiates. For example, such table can be generated by code below:

xx, yy = np.meshgrid(np.arange(100), np.arange(100))
zz = some_function(xx, yy)

现在我有一个网格上散点的坐标列表，例如，我可以通过以下方式生成这样的坐标:

now I have a list of coordinates of scattered points on the grid, for example, I can generate such coordinates by:

xs = np.random.rand(10) * 100
ys = np.random.rand(10) * 100

scattered_points = [(x, y) for x, y in zip(xs, ys)]

如何为这 10 个点插入 z 值?我只发现我可以通过搜索 (x, y) 的相邻网格点并在那里执行插值来做到这一点.有几点很好，但我只是想知道是否有更好的/pythonic 方法使用 numpy/scipy 来做到这一点.

How can I interpolate out the z value for these 10 points? I only find out that I can do this by searching the neighboring grid points of (x, y) and perform interpolation there. It is fine for a few points but I just wonder if there is a better/pythonic way to do that using numpy/scipy.

推荐答案

这应该适用于您所描述的情况.(更新)

This should work for what you describe. (Updated)

from scipy.interpolate import interp2d
import numpy as np

xx, yy = np.meshgrid(np.arange(100), np.arange(100))
zz = some_function(xx,yy)

f = interp2d(xx, yy, zz, kind='cubic')

xs = np.random.rand(10) * 100
ys = np.random.rand(10) * 100
zs = np.zeros_like(xs)

for i in range(xs.shape[0]):
    zs[i]=f(xs[i],ys[i])

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