Question

我在球体上有非常粗糙的3D测量数据，我想进行插值。我发现scipy.interpolate中的RectSphereBivariateSpline应该是最合适的。我使用RectSphereBivariateSpline文档中的示例作为起点，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.linspace(1,360,360)
phiNew = phiIndex*np.pi/180
thetaIndex = np.linspace(1,180,180)
thetaNew = thetaIndex*np.pi/180
thetaNew,phiNew = np.meshgrid(thetaNew,phiNew)
# create interpolator object and interpolate
data = r(thetaMesh,phiMesh)
lut = RectSphereBivariateSpline(theta,phi,data.T)
data_interp = lut.ev(thetaNew.ravel(),phiNew.ravel()).reshape((360,180)).T

x = (data_interp(thetaIndex,phiIndex)*np.cos(phiNew)*np.sin(thetaNew))
y = (-data_interp(thetaIndex,phiIndex)*np.sin(phiNew)*np.sin(thetaNew))
z = (data_interp(thetaIndex,phiIndex)*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()

文档中的示例有效，但是当我尝试使用以下测试数据运行上面的代码时：testdata我在声明RectSphereBivariateSpline插值器对象的代码位置得到一个ValueError：

ValueError异常：错误：输入时，输入数据的有效性受到控制必须满足以下限制。 -1＆lt; = iopt（1）＆lt; = 1,0＆lt; = iopt（2）＆lt; = 1,0＆lt; = iopt（3）＆lt; = 1， -1＆lt; = ider（1）＆lt; = 1,0＆lt; = ider（2）＆lt; = 1，ider（2）= 0，如果iopt（2）= 0。 -1＆lt; = ider（3）＆lt; = 1,0＆lt; = ider（4）＆lt; = 1，ider（4）= 0，如果iopt（3）= 0。 mu> = mumin（参见上文），mv> = 4，nuest> = 8，nvest> = 8， kwrk＆GT; = 5 +亩+ MV + nuest + nvest， lwrk＆gt; = 12 + nuest *（mv + nvest + 3）+ nvest * 24 + 4 * mu + 8 * mv + max（nuest，mv + nvest） 0℃; u（i-1）= 0：s＆gt; = 0 如果s = 0：nuest＆gt; = mu + 6 + iopt（2）+ iopt（3），nvest＆gt; = mv + 7 如果发现其中一个条件被违反，则控制权为立即重新调用到调用程序。在那种情况下没有返回近似值。

我已经尝试过，但我绝对不知道为了满足RectSphereBivariateSpline对象我应该改变什么。

有没有人暗示我可能做错了什么？

- 编辑 - 根据#HYRY的建议，我现在拥有以下运行时没有运行时错误的代码：

""" 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(1,181)
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()

然而，该图与非插值数据有很大不同，请参见图片here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）。

还有什么提示吗？

Answer 1

看起来theta[0]不能为0，如果在调用RectSphereBivariateSpline之前将其更改为litte：

theta[0] += 1e-6

SciPy RectSphereBivariateSpline在球体上插值返回ValueError

1 个答案: