我使用了在问题如何在Python中应用分段线性拟合?中找到的一些代码,以执行具有单个断点的分段线性逼近。
代码如下:
from scipy import optimize
import matplotlib.pyplot as plt
import numpy as np
%matplotlib inline
x = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ,11, 12, 13, 14, 15], dtype=float)
y = np.array([5, 7, 9, 11, 13, 15, 28.92, 42.81, 56.7, 70.59, 84.47, 98.36, 112.25, 126.14, 140.03])
def piecewise_linear(x, x0, y0, k1, k2):
return np.piecewise(x,
[x < x0],
[lambda x:k1*x + y0-k1*x0, lambda x:k2*x + y0-k2*x0])
p , e = optimize.curve_fit(piecewise_linear, x, y)
xd = np.linspace(0, 15, 100)
plt.plot(x, y, "o")
plt.plot(xd, piecewise_linear(xd, *p))
我正在尝试找出如何扩展此代码以处理n个断点。我尝试了以下代码来处理2个断点的piecewise_linear()方法,但它并没有改变任何方式中断点的值。
x = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20], dtype=float)
y = np.array([5, 7, 9, 11, 13, 15, 28.92, 42.81, 56.7, 70.59, 84.47, 98.36, 112.25, 126.14, 140.03, 150, 152, 154, 156, 158])
def piecewise_linear(x, x0, x1, a1, b1, a2, b2, a3, b3):
return np.piecewise(x,
[x < x0, np.logical_and(x >= x0, x < x1), x >= x1 ],
[lambda x:a1*x + b1, lambda x:a2*x+b2, lambda x: a3*x + b3])
p , e = optimize.curve_fit(piecewise_linear, x, y)
xd = np.linspace(0, 20, 100)
plt.plot(x, y, "o")
plt.plot(xd, piecewise_linear(xd, *p))
非常感谢您提供任何意见。