用Python进行样条插值 [英] Spline Interpolation with Python

问题描述

我编写了以下代码来执行样条插值:

I wrote the following code to perform a spline interpolation:

import numpy as np
import scipy as sp

x1 = [1., 0.88,  0.67,  0.50,  0.35,  0.27, 0.18,  0.11,  0.08,  0.04,  0.04,  0.02]
y1 = [0., 13.99, 27.99, 41.98, 55.98, 69.97, 83.97, 97.97, 111.96, 125.96, 139.95, 153.95]

x = np.array(x1)
y = np.array(y1)

new_length = 25
new_x = np.linspace(x.min(), x.max(), new_length)
new_y = sp.interpolate.interp1d(x, y, kind='cubic')(new_x)

但是我得到了:

ValueError: A value in x_new is below the interpolation range.

在interpolate.py

任何帮助将不胜感激.

推荐答案

来自有关scipy.interpolate.interp1d的科学文档:

scipy.interpolate.interp1d(x，y，kind ='linear'，axis = -1，copy = True，bounds_error = True，fill_value = np.nan)

scipy.interpolate.interp1d(x, y, kind='linear', axis=-1, copy=True, bounds_error=True, fill_value=np.nan)

x:类似array_.一维数组，其单调递增的实数值.

x : array_like. A 1-D array of monotonically increasing real values.

...

问题在于x值不是单调递增.实际上，它们正在单调减少.让我知道这是否有效，以及它是否仍在您要寻找的计算中.:

The problem is that the x values are not monotonically increasing. In fact they are monotonically decreasing. Let me know if this works and if its still the computation you are looking for.:

import numpy as np
import scipy as sp
from scipy.interpolate import interp1d

x1 = sorted([1., 0.88, 0.67, 0.50, 0.35, 0.27, 0.18, 0.11, 0.08, 0.04, 0.04, 0.02])
y1 = [0., 13.99, 27.99, 41.98, 55.98, 69.97, 83.97, 97.97, 111.96, 125.96, 139.95, 153.95]

new_length = 25
new_x = np.linspace(x.min(), x.max(), new_length)
new_y = sp.interpolate.interp1d(x, y, kind='cubic')(new_x)

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