Question

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)

但我得到了：

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

interpolate.py

中的

任何帮助都将不胜感激。

Answer 1

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。一个单调递增实数值的一维数组。

...

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，如果它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)

Answer 2

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)

Answer 3

我刚遇到上述错误，并通过删除X和Y数组中的重复值进行了修复。

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]))

使用Python进行样条插值

3 个答案: