我编写了以下代码来执行样条插值:
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)
但我得到:
ValueError: A value in x_new is below the interpolation range.
在 interpolate.py
如有任何帮助，我们将不胜感激。
最佳答案
来自scipy documentation on scipy.interpolate.interp1d :
scipy.interpolate.interp1d(x, y, kind='linear', axis=-1, copy=True, bounds_error=True, fill_value=np.nan)
x : array_like. A 1-D array of monotonically increasing real values.
...
问题是 x 值不是 monotonically increasing .事实上，它们是单调递减的。让我知道这是否有效以及它是否仍然是您正在寻找的计算。:
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)
关于python - 用 Python 进行样条插值，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/11851770/