用于在样品位置不均匀的情况下计算3D梯度的功能

问题描述

我有一个实验观察结果:

I have experimental observations in a volume:

import numpy as np

# observations are not uniformly spaced 
x = np.random.normal(0, 1, 10)
y = np.random.normal(5, 2, 10)
z = np.random.normal(10, 3, 10)
xx, yy, zz = np.meshgrid(x, y, z, indexing='ij')

# fake temperatures at those coords
tt = xx*2 + yy*2 + zz*2

# sample distances
dx = np.diff(x)
dy = np.diff(y)
dz = np.diff(z)

grad = np.gradient(tt, [dx, dy, dz])  # returns error

这给了我错误:

根据@ jay-kominek在以下评论中:

according to @jay-kominek in the comments below:

我已经更新了问题.有什么功能可以进行计算吗?

I've updated the question. Is there any function which can can do my computation?

推荐答案

注意两点:首先，标量是单个值，而不是数组.其次，函数的签名为numpy.gradient(f, *varargs, **kwargs).注意varargs前的*.这意味着如果varargs是列表，则传递*varargs.或者，您可以只提供varargs的元素作为单独的参数.

Two things to note: First, scalars are single values, not arrays. Second, the signature of the function is numpy.gradient(f, *varargs, **kwargs). Note the * before varargs. That means if varargs is a list, you pass *varargs. Or you can just provide the elements of varargs as separate arguments.

因此，np.gradient希望沿每个维度的距离为单个值，例如:

So, np.gradient wants a single value for the distance along each dimension, like:

np.gradient(tt, np.diff(x)[0], np.diff(y)[0], np.diff(z)[0])

或:

distances = [np.diff(x)[0], np.diff(y)[0], np.diff(z)[0]]
np.gradient(tt, *distances)

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