2025年python interpolate.interp1d_索引错误scipy.interpolate.interp1d「建议收藏」

编程汇总 • 2025-09-16 16:46 • 阅读 76

我试图得到一个三次样条函数scipy.interpolate.interp1d功能。我试图让documentation page上的示例正常工作，但每当我运行它时，都会出现以下错误：plt.plot(x,y,’o’,xnew,f(xnew),’-‘, xnew, f2(xnew),’–‘) File

“/Library/Python/2.7/site-packages/scipy-0.12.0.dev_ddd617d_20120920-py2.7-macosx-10.8-x86_64.egg/scipy/interpolate/interpolate.py”,

line 396, in call

y_new = self._call(x_new) File “/Library/Python/2.7/site-packages/scipy-0.12.0.dev_ddd617d_20120920-py2.7-macosx-10.8-x86_64.egg/scipy/interpolate/interpolate.py”,

line 372, in _call_spline

result = spleval(self._spline,x_new.ravel()) File “/Library/Python/2.7/site-packages/scipy-0.12.0.dev_ddd617d_20120920-py2.7-macosx-10.8-x86_64.egg/scipy/interpolate/interpolate.py”,

line 835, in spleval

res[sl] = _fitpack._bspleval(xx,xj,cvals[sl],k,deriv) IndexError: too many indices

所以，它适用于线性插值，但不适用于立方。我可能犯了一些愚蠢的错误，但我不知道出了什么问题。下面是我使用的示例代码：import numpy as np

from scipy.interpolate import interp1d

x = np.linspace(0, 10, 40)

y = np.cos(-x**2/8.0)

f = interp1d(x, y)

f2 = interp1d(x, y, kind=’cubic’)

xnew = np.linspace(0, 10, 10)

import matplotlib.pyplot as plt

plt.plot(x,y,’o’,xnew,f(xnew),’-‘, xnew, f2(xnew),’–‘)

plt.legend([‘data’, ‘linear’, ‘cubic’], loc=’best’)

plt.show()

2025年python interpolate.interp1d_索引错误scipy.interpolate.interp1d「建议收藏」

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