python - 如何在球体中生成点并用 pyplot 绘制它们？

Question

我使用这种方法来给我一个球体内点的列表。然而，当我绘制结果时，它看起来根本不是球形的。这里面的逻辑肯定有问题。可能是什么？

def give_sphere(x, y, z, r, num):
    """The distribution of dots in the sphere increases towards the center.
    Return: A List of Points (x,y,z) which are all inside the sphere."""
    points = []
    for i in range(0, num):
        factor = normedgauss()        # A value between 0 and 1 following a gaussian
        ir = r * factor
        ix = x + ir * np.cos(npi())
        iy = y + ir * np.sin(npi())
        iz = z + ir * np.cos(npi())
        points.append((ix, iy, iz))
    return points

这是 3D 绘图: 我还想在 3D 中使用 pyplot 绘制这个点列表。我可以使用以下代码来实现这一点，但随后我无法添加另一个点云以在同一图表中显示。我将如何做到这一点？

def plot_sphere(points):
    x_list = [x for [x, y, z] in points]
    y_list = [y for [x, y, z] in points]
    z_list = [z for [x, y, z] in points]

    fig = plt.figure()
    ax = Axes3D(fig)

    ax.scatter(x_list, y_list, z_list)
    plt.show()

Answer 1

您可能正在使用均匀分布的随机数生成角度，但事实并非如此。 3D 中的体积微分类似于 (dr^3)(d cos theta) (d phi)，这意味着均匀分布的变量是 cos theta，不是 theta (径向分量也是如此，但我不确定你想要做什么，所以我保持不变)

def give_sphere(x, y, z, r, num):
    points = []
    for i in range(0, num):
        factor = normedgauss()        # A value between 0 and 1 following a gaussian
        ir = r * factor
        itheta = np.arccos(np.random.uniform(-1, 1))
        iphi = np.random.uniform(0, 2 * np.pi)
        ix = x + ir * np.sin(itheta) * np.cos(iphi)
        iy = y + ir * np.sin(itheta) * np.sin(iphi)
        iz = z + ir * np.cos(itheta)
        points.append((ix, iy, iz))
    return points

考虑到这一点，这就是您应该得到的

至于第二个问题

def plot_sphere(points, ax):
    x_list = [x for [x, y, z] in points]
    y_list = [y for [x, y, z] in points]
    z_list = [z for [x, y, z] in points]

    ax.scatter(x_list, y_list, z_list)


fig = plt.figure()
ax = Axes3D(fig)


points1 = give_sphere(0, 0, -2, 1, 1000)
points2 = give_sphere(0, 0, 2, 1, 1000)
plot_sphere(points1, ax)
plot_sphere(points2, ax)

plt.show()