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

Question

我有一个非常粗糙的球体上的3D测量数据，我想插入。有了@ M4rtini和@HYRY在stackoverflow的帮助，我现在已经能够生成工作代码（基于来自SciPy的RectSphereBivariateSpline示例的原始示例）。

测试数据可以在这里找到：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() 

虽然代码运行时，所产生的情节是不是非插数据太大的不同，看到的画面

为一个参考。

而且，在运行交互式会话时，data_interp的值（> 3e5）比原始数据（大约20 max）大得多。

有没有人知道我可能做错了什么？

Answer 1

我似乎已经解决了它！

对于事物，我试图外推，而我只能插入这些分散的数据。所以新的插值网格应该只能达到θ= 140度左右。

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

所以现在我有以下代码：

""" 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() 

所得的情节很好地比较原始的非插数据：

我不完全理解为什么S的关系被设置为900，因为RectSphereBivariateSpline文档说插值的s = 0。但是，在阅读文档时，我会进一步了解一些内容：

选择s的最佳值可能是一项棘手的任务。 s的推荐值取决于数据值的准确性。如果用户对数据有统计错误的想法，她也可以为s找到适当的估计值。通过假设，如果她指定了正确的s，内插器将使用精确地再现数据底层函数的样条函数f（u，v），她可以评估sum（（r（i，j）-s（u ），v（j）））** 2）为这个s找到一个好的估计。例如，如果她知道她的r（i，j）值的统计误差不大于0.1，她可能会认为一个好的s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。最好的方法是从一个非常大的s值开始（以确定最小二乘多项式和s的相应上界fp0），然后逐渐减小s的值（例如在开始时减少10倍，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。