[英]Interpolating non-uniformly distributed points on a 3D sphere
我在單位球面上有幾個點,它們根據https://www.cmu.edu/biolphys/deserno/pdf/sphere_equi.pdf 中描述的算法分布(並在下面的代碼中實現)。 在這些點中的每一個上,我都有一個值,在我的特定情況下,它表示 1 減去一個小錯誤。 如果這很重要,錯誤在[0, 0.1]
,所以我的值在[0.9, 1]
。
可悲的是,計算錯誤是一個代價高昂的過程,我無法根據需要計算盡可能多的點。 不過,我希望我的情節看起來像我在繪制“連續”的東西。 所以我想為我的數據擬合一個插值函數,以便能夠根據需要采樣盡可能多的點。
經過一些研究,我發現scipy.interpolate.SmoothSphereBivariateSpline似乎完全符合我的要求。 但我不能讓它正常工作。
問題:我可以用什么來插值(樣條、線性插值,目前什么都可以)我在單位球體上的數據? 答案可以是“您誤用了scipy.interpolation
,這是執行此操作的正確方法”或“此其他功能更適合您的問題”。
安裝numpy
和scipy
應該可執行的示例代碼:
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)
上面代碼的問題:
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
實例時引發。s
值。 我測試了s
不同值,范圍從0.0001
到100000
,上面的代碼總是引發,上述異常或: ValueError: Error code returned by bispev: 10
編輯:我在這里包括我的發現。 它們不能真正被視為解決方案,這就是為什么我正在編輯而不是作為答案發布。
通過更多的研究,我發現了這個問題Using Radial Basis Functions to Interpolate a Function on a Sphere 。 作者和我有完全相同的問題,並使用了不同的插值器: scipy.interpolate.Rbf 。 我通過替換插值器和繪圖更改了上面的代碼:
# 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()
除了在不連續的一行上,插值似乎還可以,就像在引導我上這門課的問題中一樣。 答案之一給出了使用不同距離的想法,更適合球面坐標:Haversine 距離。
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)
執行時會發出警告:
LinAlgWarning: Ill-conditioned matrix (rcond=1.33262e-19): result may not be accurate.
self.nodes = linalg.solve(self.A, self.di)
並且結果完全不是預期的:插值函數的值可能會上升到-1
,這顯然是錯誤的。
您可以使用笛卡爾坐標而不是球坐標。
Rbf使用的默認范數參數 ( 'euclidean'
) 就足夠了
# interpolation
x, y, z = numpy.array(points).T
interpolator = scipy.interpolate.Rbf(x, y, z, r_interp)
# predict
heatmap = interpolator(X, Y, Z)
結果如下:
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,您還可以使用余弦距離(范數參數):
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)
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