scipy.interp2d 警告和与预期不同的结果

Question

我正在尝试将 MATLAB 代码转换为等效的 python。 我有 3 个数组，我想计算 interp2d:

nuA = np.asarray([2.439,2.5,2.6,2.7,2.8,3.0,3.2,3.5,4.0,5.0,6.0,8.0,10,15,25])
nuB = np.asarray([0,0.1,0.2,0.3,0.5,0.7,1])
a, b = np.meshgrid(nuA, nuB)
betaTab  = np.transpose(np.asarray([[0.0,2.16,1.0,1.0,1.0,1.0,1.0],[0.0,1.592,3.39,1.0,1.0,1.0,1.0],[0.0,0.759,1.8,1.0,1.0,1.0,1.0],[0.0,0.482,1.048,1.694,1.0,1.0,1.0],[0.0,0.36,0.76,1.232,2.229,1.0,1.0],[0.0,0.253,0.518,0.823,1.575,1.0,1.0],[0.0,0.203,0.41,0.632,1.244,1.906,1.0],[0.0,0.165,0.332,0.499,0.943,1.56,1.0],[0.0,0.136,0.271,0.404,0.689,1.23,2.195],[0.0,0.109,0.216,0.323,0.539,0.827,1.917],[0.0,0.096,0.19,0.284,0.472,0.693,1.759],[0.0,0.082,0.163,0.243,0.412,0.601,1.596],[0.0,0.074,0.147,0.22,0.377,0.546,1.482],[0.0,0.064,0.128,0.191,0.33,0.478,1.362],[0.0,0.056,0.112,0.167,0.285,0.428,1.274]]))
ip = scipy.interpolate.interp2d(a,b,betaTab)

当我尝试运行它时，会显示以下警告：

/usr/local/lib/python2.7/dist-packages/scipy/interpolate/fitpack.py:981: RuntimeWarning: No more knots can be added because the additional knot would
coincide with an old one. Probable cause: s too small or too large
a weight to an inaccurate data point. (fp>s)
    kx,ky=1,1 nx,ny=4,14 m=105 fp=21.576347 s=0.000000
  warnings.warn(RuntimeWarning(_iermess2[ierm][0] + _mess))

我知道interp2d 与matlab interp2 不同，在python 中RectBivariateSpline 函数是首选。但由于我的数据长度，我不能使用后一个函数。此外，ip(xi,yi) 的最终结果与 MATLAB 答案不同。

如何在没有警告的情况下计算 interp2d 并正确计算？

Answer 1

您的输入数据似乎定义不明确。这是您的输入点的表面图：

这不是一个容易插值的问题。顺便说一句，我有recently ran into problems interp2d 甚至无法插入平滑的数据集。所以我建议改为查看scipy.interpolate.griddata：

import numpy as np
import scipy.interpolate as interp
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

#define your data as you did in your question: a, b and betaTab

ip = interp.interp2d(a,b,betaTab)           # original interpolator

aplotv = np.linspace(a.min(),a.max(),100)   # to interpolate at
bplotv = np.linspace(b.min(),b.max(),100)   # to interpolate at
aplot,bplot = np.meshgrid(aplotv,bplotv)    # mesh to interpolate at

# actual values from interp2d:
betainterp2d = ip(aplotv,bplotv)

# actual values from griddata:
betagriddata = interp.griddata(np.array([a.ravel(),b.ravel()]).T,betaTab.ravel(),np.array([aplot.ravel(),bplot.ravel()]).T)
# ^ this probably could be written in a less messy way,
# I'll keep thinking about it

#plot results
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_surface(aplot,bplot,betainterp2d,cmap='viridis',cstride=1,rstride=1)

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot_surface(aplot,bplot,betagriddata,cmap='viridis',cstride=1,rstride=1)

结果：（左：interp2d，右：griddata）

结论：使用scipy.interpolate.griddata。