我想对以下数据点进行曲线拟合:
import matplotlib.pyplot as plt
import numpy as np
from scipy.optimize import curve_fit
t = np.array([15474.6, 15475.6, 15476.6, 15477.6, 15478.6, 15479.6, 15480.6,
15481.6, 15482.6, 15483.6, 15484.6, 15485.6, 15486.6, 15487.6,
15488.6, 15489.6, 15490.6, 15491.6, 15492.6, 15493.6, 15494.6,
15495.6, 15496.6, 15497.6, 15498.6, 15499.6, 15500.6, 15501.6,
15502.6, 15503.6, 15504.6, 15505.6, 15506.6, 15507.6, 15508.6,
15509.6, 15510.6, 15511.6, 15512.6, 15513.6])
v = np.array([4.082, 4.133, 4.136, 4.138, 4.139, 4.14, 4.141, 4.142, 4.143,
4.144, 4.144, 4.145, 4.145, 4.147, 4.146, 4.147, 4.148, 4.148,
4.149, 4.149, 4.149, 4.15, 4.15, 4.15, 4.151, 4.151, 4.152,
4.152, 4.152, 4.153, 4.153, 4.153, 4.153, 4.154, 4.154, 4.154,
4.154, 4.154, 4.155, 4.155])
我想要拟合数据的指数函数是:
表示上述公式的 Python 函数以及与数据拟合的相关曲线详述如下:
def func(t, a, b, alpha):
return a - b * np.exp(-alpha * t)
# scale vector to start at zero otherwise exponent is too large
t_scale = t - t[0]
# initial guess for curve fit coefficients
a0 = v[-1]
b0 = v[0]
alpha0 = 1/t_scale[-1]
# coefficients and curve fit for curve
popt4, pcov4 = curve_fit(func, t_scale, v, p0=(a0, b0, alpha0))
a, b, alpha = popt4
v_fit = func(t_scale, a, b, alpha)
ss_res = np.sum((v - v_fit) ** 2) # residual sum of squares
ss_tot = np.sum((v - np.mean(v)) ** 2) # total sum of squares
r2 = 1 - (ss_res / ss_tot) # R squared fit, R^2
与曲线拟合相比的数据绘制如下。还提供了参数和 R 平方值。
a0 = 4.1550 b0 = 4.0820 alpha0 = 0.0256
a = 4.1490 b = 0.0645 alpha = 0.9246
R² = 0.8473
是否可以使用上述方法更好地拟合数据，或者我是否需要使用不同形式的指数方程？
我也不确定使用什么作为初始值(a0、b0、alpha0)。在示例中，我从数据中选择了点，但这可能不是最好的方法。对于曲线拟合系数的初始猜测，有什么建议吗？
请您参考如下方法:
对我来说，这看起来更适合多个组件，而不是单个指数。
def func(t, a, b, c, d, e):
return a*np.exp(-t/b) + c*np.exp(-t/d) + e
# scale vector to start at zero otherwise exponent is too large
t_scale = t - t[0]
# initial guess for curve fit coefficients
guess = [1, 1, 1, 1, 0]
# coefficients and curve fit for curve
popt, pcov = curve_fit(func, t_scale, v, p0=guess)
v_fit = func(t_scale, *popt)
