Python SciPy RectSphereBivariateSpline插值生成错误的数据？

Question

我在一个非常粗糙的球体上有3D测量数据，我想进行插值。借助@ M4rtini和@HYRY在stackoverflow的帮助，我现在能够生成工作代码（基于SciPy的RectSphereBivariateSpline示例中的原始示例）。

可以在此处找到测试数据：testdata

""" read csv input file, post process and plot 3D data """
import csv
import numpy as np
from mayavi import mlab
from scipy.interpolate import RectSphereBivariateSpline

# user input
nElevationPoints = 17 # needs to correspond with csv file
nAzimuthPoints = 40 # needs to correspond with csv file
threshold = - 40 # needs to correspond with how measurement data was captured
turnTableStepSize = 72 # needs to correspond with measurement settings
resolution = 0.125 # needs to correspond with measurement settings

# read data from file
patternData = np.empty([nElevationPoints, nAzimuthPoints]) # empty buffer
ifile  = open('ttest.csv') # need the 'b' suffix to prevent blank rows being inserted
reader = csv.reader(ifile,delimiter=',')
reader.next() # skip first line in csv file as this is only text
for nElevation in range (0,nElevationPoints):
    # azimuth
    for nAzimuth in range(0,nAzimuthPoints):  
        patternData[nElevation,nAzimuth] = reader.next()[2]
ifile.close()

# post process
def r(thetaIndex,phiIndex):
    """r(thetaIndex,phiIndex): function in 3D plotting to return positive vector length from patternData[theta,phi]"""
    radius = -threshold + patternData[thetaIndex,phiIndex]
    return radius

#phi,theta = np.mgrid[0:nAzimuthPoints,0:nElevationPoints]
theta = np.arange(0,nElevationPoints)
phi = np.arange(0,nAzimuthPoints)
thetaMesh, phiMesh = np.meshgrid(theta,phi)
stepSizeRad = turnTableStepSize * resolution * np.pi / 180
theta = theta * stepSizeRad
phi = phi * stepSizeRad

# create new grid to interpolate on
phiIndex = np.arange(1,361)
phiNew = phiIndex*np.pi/180
thetaIndex = np.arange(1,181)
thetaNew = thetaIndex*np.pi/180
thetaNew,phiNew = np.meshgrid(thetaNew,phiNew)
# create interpolator object and interpolate
data = r(thetaMesh,phiMesh)
theta[0] += 1e-6 # zero values for theta cause program to halt; phi makes no sense at theta=0
lut = RectSphereBivariateSpline(theta,phi,data.T)
data_interp = lut.ev(thetaNew.ravel(),phiNew.ravel()).reshape((360,180)).T

def rInterp(theta,phi):
    """rInterp(theta,phi): function in 3D plotting to return positive vector length from interpolated patternData[theta,phi]"""
    thetaIndex = theta/(np.pi/180)
    thetaIndex = thetaIndex.astype(int)
    phiIndex = phi/(np.pi/180)
    phiIndex = phiIndex.astype(int)
    radius = data_interp[thetaIndex,phiIndex]
    return radius
# recreate mesh minus one, needed otherwise the below gives index error, but why??
phiIndex = np.arange(0,360)
phiNew = phiIndex*np.pi/180
thetaIndex = np.arange(0,180)
thetaNew = thetaIndex*np.pi/180
thetaNew,phiNew = np.meshgrid(thetaNew,phiNew)

x = (rInterp(thetaNew,phiNew)*np.cos(phiNew)*np.sin(thetaNew))
y = (-rInterp(thetaNew,phiNew)*np.sin(phiNew)*np.sin(thetaNew))
z = (rInterp(thetaNew,phiNew)*np.cos(thetaNew))

# plot 3D data
obj = mlab.mesh(x, y, z, colormap='jet')
obj.enable_contours = True
obj.contour.filled_contours = True
obj.contour.number_of_contours = 20
mlab.show()

虽然代码运行，但结果图与非内插数据有很大不同，见图片

作为参考。

此外，在运行交互式会话时，data_interp的值（> 3e5）远大于原始数据（大约20个最大值）。

有没有人知道我可能做错了什么？

Answer 1

我好像已经解决了！

对于事情，我试图推断，而我只能插入这些分散的数据。因此，新的插值网格应该只达到θ= 140度左右。

但最重要的变化是在RectSphereBivariateSpline调用中添加参数s = 900。

所以我现在有以下代码：

""" read csv input file, post process and plot 3D data """
import csv
import numpy as np
from mayavi import mlab
from scipy.interpolate import RectSphereBivariateSpline

# user input
nElevationPoints = 17 # needs to correspond with csv file
nAzimuthPoints = 40 # needs to correspond with csv file
threshold = - 40 # needs to correspond with how measurement data was captured
turnTableStepSize = 72 # needs to correspond with measurement settings
resolution = 0.125 # needs to correspond with measurement settings

# read data from file
patternData = np.empty([nElevationPoints, nAzimuthPoints]) # empty buffer
ifile  = open('ttest.csv') # need the 'b' suffix to prevent blank rows being inserted
reader = csv.reader(ifile,delimiter=',')
reader.next() # skip first line in csv file as this is only text
for nElevation in range (0,nElevationPoints):
    # azimuth
    for nAzimuth in range(0,nAzimuthPoints):  
        patternData[nElevation,nAzimuth] = reader.next()[2]
ifile.close()

# post process
def r(thetaIndex,phiIndex):
    """r(thetaIndex,phiIndex): function in 3D plotting to return positive vector length from patternData[theta,phi]"""
    radius = -threshold + patternData[thetaIndex,phiIndex]
    return radius

#phi,theta = np.mgrid[0:nAzimuthPoints,0:nElevationPoints]
theta = np.arange(0,nElevationPoints)
phi = np.arange(0,nAzimuthPoints)
thetaMesh, phiMesh = np.meshgrid(theta,phi)
stepSizeRad = turnTableStepSize * resolution * np.pi / 180
theta = theta * stepSizeRad
phi = phi * stepSizeRad

# create new grid to interpolate on
phiIndex = np.arange(1,361)
phiNew = phiIndex*np.pi/180
thetaIndex = np.arange(1,141)
thetaNew = thetaIndex*np.pi/180
thetaNew,phiNew = np.meshgrid(thetaNew,phiNew)
# create interpolator object and interpolate
data = r(thetaMesh,phiMesh)
theta[0] += 1e-6 # zero values for theta cause program to halt; phi makes no sense at theta=0
lut = RectSphereBivariateSpline(theta,phi,data.T,s=900)
data_interp = lut.ev(thetaNew.ravel(),phiNew.ravel()).reshape((360,140)).T

def rInterp(theta,phi):
    """rInterp(theta,phi): function in 3D plotting to return positive vector length from interpolated patternData[theta,phi]"""
    thetaIndex = theta/(np.pi/180)
    thetaIndex = thetaIndex.astype(int)
    phiIndex = phi/(np.pi/180)
    phiIndex = phiIndex.astype(int)
    radius = data_interp[thetaIndex,phiIndex]
    return radius
# recreate mesh minus one, needed otherwise the below gives index error, but why??
phiIndex = np.arange(0,360)
phiNew = phiIndex*np.pi/180
thetaIndex = np.arange(0,140)
thetaNew = thetaIndex*np.pi/180
thetaNew,phiNew = np.meshgrid(thetaNew,phiNew)

x = (rInterp(thetaNew,phiNew)*np.cos(phiNew)*np.sin(thetaNew))
y = (-rInterp(thetaNew,phiNew)*np.sin(phiNew)*np.sin(thetaNew))
z = (rInterp(thetaNew,phiNew)*np.cos(thetaNew))

# plot 3D data
intensity = rInterp(thetaNew,phiNew)
obj = mlab.mesh(x, y, z, scalars = intensity, colormap='jet')
obj.enable_contours = True
obj.contour.filled_contours = True
obj.contour.number_of_contours = 20
mlab.show()

得到的图与原始的非插值数据很好地比较：

我不完全理解为什么s应设置为900，因为RectSphereBivariateSpline文档说s = 0用于插值。但是，在进一步阅读文档时，我获得了一些见解：

  

选择s的最佳值可能是一项微妙的任务。 s的推荐值取决于数据值的准确性。如果用户知道数据的统计错误，她也可以找到s的正确估计。通过假设，如果她指定了正确的s，插值器将使用精确再现数据基础函数的样条f（u，v），她可以求和（（r（i，j）-s（u（i） ），v（j）））** 2）找到一个很好的估计s。例如，如果她知道她的r（i，j）值上的统计误差不大于0.1，她可能会认为商品的值应该不大于u.size * v.size *（0.1 ）** 2。   如果对r（i，j）中的统计误差一无所知，则s必须通过反复试验来确定。然后最好从一个非常大的s值（确定最小二乘多项式和s的相应上限fp0）开始，然后逐步减小s的值（比如开始时的因子10，即s = fp0 / 10，fp0 / 100，......以及更仔细的近似显示更多细节）以获得更接近的拟合。