用于在球体表面均匀分布点的万无一失的算法？

Question

我一直试图在半径为“inner_radius”的球体表面上生成点，以便它们均匀分布。该算法在半径为 1 时按预期工作，但在半径较大时生成的点少于预期。 我在这里查看了类似的问题，但它们似乎是用于在整个体积中生成点，而不仅仅是在球体的表面上。

import numpy as np
PI=np.pi

def spherical_to_cartesian(pol_ang,azim_ang,radius): #This function converts given spherical coordinates (theta, phi and radius) to cartesian coordinates.
    return np.array((radius*np.sin(pol_ang) * np.cos(azim_ang),
                        radius*np.sin(pol_ang) * np.sin(azim_ang),
                        radius*np.cos(pol_ang))
                        )

def get_electron_coordinates_list(inner_radius,electron_count):
    #Algorithm used was mostly  taken from https://www.cmu.edu/biolphys/deserno/pdf/sphere_equi.pdf . Explanations in code added by me.
    electron_coordinate_list=[]
    inner_area=4*(PI*inner_radius**2)
    area_per_electron=inner_area/electron_count
    pseudo_length_per_electron=np.sqrt(area_per_electron) #This is the side length of a square where the area of it is the area per electron on the sphere.
    #Now, we need to get a value of angular space, such that angular space between electrons on latitude and longitude per electron is equal
    #As a first step to obtaining this, we must make another value holding a whole number approximation of the ratio between PI and the pseudo_length. This will give the number of 
    #possible latitudes.

    possible_count_of_lats=np.round(PI/pseudo_length_per_electron)

    approx_length_per_electron_lat=PI/possible_count_of_lats #This is the length between electrons on a latitude
    approx_length_per_electron_long=area_per_electron/approx_length_per_electron_lat #This is the length between electrons on a longitude

    for electron_num_lat in range(int(possible_count_of_lats.item())): #The int(somenumpyvalue.item()) is used because Python cannot iterate over a numpy integer and it must be converted to normal int.
        pol_ang=PI*(electron_num_lat+0.5)/possible_count_of_lats #The original algorithm recommended pol_ang=PI*(electron_num_lat+0.5)/possible_count_of_lats. The 0.5 appears to be added in order to get a larger number of coordinates.
        #not sure if removing the 0.5 affects results. It didnt do so drastically, so what gives? Anyway, this gets the polar angle as PI*(latitudenumber)/totalnumberoflatitudes.

        possible_count_of_longs=np.round(2*PI*np.sin(pol_ang)/approx_length_per_electron_long)

        for electron_num_long in range(int(possible_count_of_longs.item())):

            azim_ang=(2*PI)*(electron_num_long)/possible_count_of_longs #This gets the azimuthal angle as 2PI*longitudenumber/totalnumberoflongitudes

            electron_coordinate=spherical_to_cartesian(pol_ang, azim_ang,inner_radius) #Converts the recieved spherical coordinates to cartesian so Manim can easily handle them.
            electron_coordinate_list.append(electron_coordinate) #Add this coordinate to the electron_coordinate_list

            print("Got coordinates: ",electron_coordinate) #Print the coordinate recieved.
    print(len(electron_coordinate_list)," points generated.") #Print the amount of electrons will exist. Comment these two lines out if you don't need the data.

    return electron_coordinate_list 
get_electron_coordinates_list(1,100)
get_electron_coordinates_list(2,100)

Spherical_to_Cartesian() 只是将球面点转换为笛卡尔坐标。

对于 100 个点和半径 1，它会生成 99 个点。 但是，如果半径为 2 并且请求 100 个点，则只生成 26 个点。

Answer 1

如果您可以在球体的体积中均匀地生成点，那么要在球体的表面上获得均匀的分布，您可以简单地对向量进行归一化，使其半径等于球体的半径。

或者，您可以使用独立同分布正态分布为rotationally-invariant这一事实。如果您从平均值为 1 且标准差为 0 的 3 个正态分布中采样，然后同样对向量进行归一化，则它将在球体表面上是均匀的。这是一个例子：

import random

def sample_sphere_surface(radius=1):
    x, y, z = (random.normalvariate(0, 1) for i in range(3))
    scalar = radius / (x**2 + y**2 + z**2) ** 0.5
    return (x * scalar, y * scalar, z * scalar)

为了绝对万无一失，当x、y 和z 都恰好为零时，我们可以处理天文数字上不太可能出现的除零错误： p>

def sample_sphere_surface(radius=1):
    while True:
        try:
            x, y, z = (random.normalvariate(0, 1) for i in range(3))
            scalar = radius / (x**2 + y**2 + z**2) ** 0.5
            return (x * scalar, y * scalar, z * scalar)
        except ZeroDivisionError:
            pass

Answer 2

区域的元素在极坐标中是sinΘ dΘ dφ。因此方位角可以均匀分布，而倾角必须重新分布。使用逆变换采样技巧，Θ=arccos(u) 其中u 被统一绘制就可以了。

因此在笛卡尔坐标中，(√(1-u²) cos v, √(1-u²) sin v, u)，其中u 来自[-1,1)，v 来自[0,2π)。