样条插值

Question

我在对以下集合执行样条插值时遇到困难：

import numpy
SOURCE = numpy.array([[1,2,3],[3,4,5], [9,10,11]])
from scipy.interpolate import griddata
from scipy.interpolate import interp1d
input = [0.5,2,3,6,9,15]

线性插值工作正常，但是当我用三次方替换线性时，出现错误:

f = interp1d(SOURCE[:,0], SOURCE[:,1:], kind="linear", axis=0, bounds_error=False)
f(input)

f = interp1d(SOURCE[:,0], SOURCE[:,1:], kind="cubic", axis=0, bounds_error=False)
ValueError: The number of derivatives at boundaries does not match: expected 1, got 0+0

如何执行此三次插值？

Answer 1

您的 SOURCE 数据太短。三次样条至少需要四个点进行插值，但您只提供了三个。如果您向 SOURCE 添加一个值，它应该或多或少按预期工作：

>>> SOURCE = numpy.array([[1,2,3],[3,4,5], [9,10,11], [12,13,14]])  # added an extra value
>>> f = interp1d(SOURCE[:,0], SOURCE[:,1:], kind="cubic", axis=0, bounds_error=False)
>>> f(input)
array([[nan, nan],
       [ 3.,  4.],
       [ 4.,  5.],
       [ 7.,  8.],
       [10., 11.],
       [nan, nan]])

样条插值

Spline interpolation

python

interpolation

scipy

cubic-spline