python - 在 3D 球体上插入非均匀分布的点

Question

我在单位球面上有几个点根据 https://www.cmu.edu/biolphys/deserno/pdf/sphere_equi.pdf 中描述的算法分布(并在下面的代码中实现)。在这些点中的每一个上，我都有一个值，在我的特定情况下，它表示 1 减去一个小错误。错误在 [0, 0.1]如果这很重要，那么我的值在 [0.9, 1] .
可悲的是，计算错误是一个代价高昂的过程，我无法根据需要计算尽可能多的点。不过，我希望我的情节看起来像我在绘制“连续”的东西。
所以我想为我的数据拟合一个插值函数，以便能够根据需要采样尽可能多的点。
经过一番研究，我发现 scipy.interpolate.SmoothSphereBivariateSpline这似乎正是我想要的。但我不能让它正常工作。
问题:我可以用什么来插值(样条，线性插值，目前什么都可以)我在单位球体上的数据？答案可以是“您误用了 scipy.interpolation ，这是执行此操作的正确方法”或“此其他功能更适合您的问题”。
应该可以使用 numpy 执行的示例代码和 scipy安装:

import typing as ty

import numpy
import scipy.interpolate


def get_equidistant_points(N: int) -> ty.List[numpy.ndarray]:
    """Generate approximately n points evenly distributed accros the 3-d sphere.

    This function tries to find approximately n points (might be a little less
    or more) that are evenly distributed accros the 3-dimensional unit sphere.

    The algorithm used is described in
    https://www.cmu.edu/biolphys/deserno/pdf/sphere_equi.pdf.
    """
    # Unit sphere
    r = 1

    points: ty.List[numpy.ndarray] = list()

    a = 4 * numpy.pi * r ** 2 / N
    d = numpy.sqrt(a)
    m_v = int(numpy.round(numpy.pi / d))
    d_v = numpy.pi / m_v
    d_phi = a / d_v

    for m in range(m_v):
        v = numpy.pi * (m + 0.5) / m_v
        m_phi = int(numpy.round(2 * numpy.pi * numpy.sin(v) / d_phi))
        for n in range(m_phi):
            phi = 2 * numpy.pi * n / m_phi
            points.append(
                numpy.array(
                    [
                        numpy.sin(v) * numpy.cos(phi),
                        numpy.sin(v) * numpy.sin(phi),
                        numpy.cos(v),
                    ]
                )
            )
    return points


def cartesian2spherical(x: float, y: float, z: float) -> numpy.ndarray:
    r = numpy.linalg.norm([x, y, z])
    theta = numpy.arccos(z / r)
    phi = numpy.arctan2(y, x)
    return numpy.array([r, theta, phi])


n = 100
points = get_equidistant_points(n)
# Random here, but costly in real life.
errors = numpy.random.rand(len(points)) / 10

# Change everything to spherical to use the interpolator from scipy.
ideal_spherical_points = numpy.array([cartesian2spherical(*point) for point in points])
r_interp = 1 - errors
theta_interp = ideal_spherical_points[:, 1]
phi_interp = ideal_spherical_points[:, 2]
# Change phi coordinate from [-pi, pi] to [0, 2pi] to please scipy.
phi_interp[phi_interp < 0] += 2 * numpy.pi

# Create the interpolator.
interpolator = scipy.interpolate.SmoothSphereBivariateSpline(
    theta_interp, phi_interp, r_interp
)

# Creating the finer theta and phi values for the final plot
theta = numpy.linspace(0, numpy.pi, 100, endpoint=True)
phi = numpy.linspace(0, numpy.pi * 2, 100, endpoint=True)

# Creating the coordinate grid for the unit sphere.
X = numpy.outer(numpy.sin(theta), numpy.cos(phi))
Y = numpy.outer(numpy.sin(theta), numpy.sin(phi))
Z = numpy.outer(numpy.cos(theta), numpy.ones(100))

thetas, phis = numpy.meshgrid(theta, phi)
heatmap = interpolator(thetas, phis)

上面代码的问题:

使用原样的代码，我有一个

ValueError: The required storage space exceeds the available storage space: nxest or nyest too small, or s too small. The weighted least-squares spline corresponds to the current set of knots.

在初始化 interpolator 时引发实例。

上面的问题好像说我应该改变s的值那是 scipy.interpolate.SmoothSphereBivariateSpline 的参数之一.我测试了 s 的不同值范围从 0.0001至 100000 , 上面的代码总是引发上述异常或:

ValueError: Error code returned by bispev: 10

编辑:我在这里包括我的发现。它们不能真正被视为解决方案，这就是为什么我正在编辑而不是作为答案发布。
通过更多的研究，我发现了这个问题 Using Radial Basis Functions to Interpolate a Function on a Sphere .作者和我有完全一样的问题，使用了不同的插值器:scipy.interpolate.Rbf .我通过替换插值器和绘图更改了上面的代码:

# Create the interpolator.
interpolator = scipy.interpolate.Rbf(theta_interp, phi_interp, r_interp)

# Creating the finer theta and phi values for the final plot
plot_points = 100
theta = numpy.linspace(0, numpy.pi, plot_points, endpoint=True)
phi = numpy.linspace(0, numpy.pi * 2, plot_points, endpoint=True)

# Creating the coordinate grid for the unit sphere.
X = numpy.outer(numpy.sin(theta), numpy.cos(phi))
Y = numpy.outer(numpy.sin(theta), numpy.sin(phi))
Z = numpy.outer(numpy.cos(theta), numpy.ones(plot_points))

thetas, phis = numpy.meshgrid(theta, phi)
heatmap = interpolator(thetas, phis)


import matplotlib as mpl
import matplotlib.pyplot as plt
from matplotlib import cm

colormap = cm.inferno
normaliser = mpl.colors.Normalize(vmin=numpy.min(heatmap), vmax=1)
scalar_mappable = cm.ScalarMappable(cmap=colormap, norm=normaliser)
scalar_mappable.set_array([])

fig = plt.figure()
ax = fig.add_subplot(111, projection="3d")
ax.plot_surface(
    X,
    Y,
    Z,
    facecolors=colormap(normaliser(heatmap)),
    alpha=0.7,
    cmap=colormap,
)
plt.colorbar(scalar_mappable)
plt.show()

此代码运行顺利并给出以下结果:

插值似乎没问题 除了 在一条不连续的线上，就像在引导我上这门课的问题中一样。 One of the answer给出使用不同距离的想法，更适应球面坐标:Haversine 距离。

def haversine(x1, x2):
    theta1, phi1 = x1
    theta2, phi2 = x2
    return 2 * numpy.arcsin(
        numpy.sqrt(
            numpy.sin((theta2 - theta1) / 2) ** 2
            + numpy.cos(theta1) * numpy.cos(theta2) * numpy.sin((phi2 - phi1) / 2) ** 2
        )
    )


# Create the interpolator.
interpolator = scipy.interpolate.Rbf(theta_interp, phi_interp, r_interp, norm=haversine)

执行时会发出警告:

LinAlgWarning: Ill-conditioned matrix (rcond=1.33262e-19): result may not be accurate.
  self.nodes = linalg.solve(self.A, self.di)

并且结果完全不是预期的:插值函数的值可能高达 -1这显然是错误的。

Answer 1

您可以使用 笛卡尔坐标而不是球坐标。
Rbf 使用的默认范数参数 ( 'euclidean' )足够了

# interpolation
x, y, z = numpy.array(points).T
interpolator = scipy.interpolate.Rbf(x, y, z, r_interp)

# predict
heatmap = interpolator(X, Y, Z)

结果如下:

ax.plot_surface(
    X, Y, Z,
    rstride=1, cstride=1, 
    # or rcount=50, ccount=50,
    facecolors=colormap(normaliser(heatmap)),
    cmap=colormap,
    alpha=0.7, shade=False
)
ax.set_xlabel('x axis')
ax.set_ylabel('y axis')
ax.set_zlabel('z axis')

如果需要，您还可以使用余弦距离(范数参数):

def cosine(XA, XB):
    if XA.ndim == 1:
        XA = numpy.expand_dims(XA, axis=0)
    if XB.ndim == 1:
        XB = numpy.expand_dims(XB, axis=0)
    return scipy.spatial.distance.cosine(XA, XB)

为了更好地看到差异，
我堆叠了两个图像，减去它们并反转图层。