Python SciPy RectSphereVariableSphere球体上的样条插值返回值错误

我有一个非常粗糙的球体上的三维测量数据，我想插值。 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()

文档中的示例是有效的，但当我尝试使用以下测试数据运行上述代码时：在声明RectSphereBivariateSpline插值器对象的代码位置，我得到一个ValueError：

值错误： 错误：输入时，输入数据受有效性控制 必须满足以下限制条件。

---

-1如果在调用

RectSphereBivariateSpline

之前稍微更改它，则看起来

theta[0]

不能为0：

theta[0] += 1e-6

---

完美的这解决了插值器对象调用！我现在看到我还有一些其他问题需要解决，但希望我能自己解决这些问题。你能告诉我你是怎么知道的吗（这样我才能变得更加自给自足）？@user3116919，我将你的数据与示例数据进行了比较，唯一不同的是你的lats数据包含零。当lat为0时，lons将没有办法，所以我认为这就是问题所在。在实现了您的解决方案并解决了之后出现的一些其他问题之后，我现在有了运行代码。请参见上面的编辑。虽然它运行，但插值数据的值与原始数据（以及结果图）相差很大。有什么想法吗？@user3116919因为stackoverflow只关注一个问题+答案，所以最好发布一个新的、孤立的、具体的问题，而不是在这个问题中展开。祝你好运看起来很有趣。谢谢，@kobejohn，我刚刚按照你的建议，把下面的问题作为一个单独的主题发布了。

theta[0] += 1e-6