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
任何帮助，将不胜感激。
从scipy.interpolate.interp1d 上的 scipy 文档：
scipy.interpolate.interp1d(x, y, kind='linear', axis=-1, copy=True, bounds_error=True, fill_value=np.nan)
x：array_like。单调递增实数值的一维数组。
...
问题是 x 值不是单调递增的。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)
您可以通过以下方式获得此信息：
import numpy as np
import scipy as sp
from scipy.interpolate import interp1d
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]
# Combine lists into list of tuples
points = zip(x1, y1)
# Sort list of tuples by x-value
points = sorted(points, key=lambda point: point[0])
# Split list of tuples into two list of x values any y values
x1, y1 = zip(*points)
new_length = 25
new_x = np.linspace(min(x1), max(x1), new_length)
new_y = sp.interpolate.interp1d(x1, y1, kind='cubic')(new_x)
我刚刚收到上述错误并通过删除 X 和 Y 数组中的重复值来修复它。
x = np.sort(np.array([0, .2, .2, .4, .6, .9]))
y = np.sort(np.sort(np.array([0, .1, .06, .11, .25, .55]))
⬇将0.2更改为0.3或任何数字。
x = np.sort(np.array([0, .2, .3, .4, .6, .9]))
y = np.sort(np.sort(np.array([0, .1, .06, .11, .25, .55]))