非线性最小二乘拟合不会最小化

Question

我正在尝试用三个指数的和来拟合衰变，并使用下面的代码。从表面上看，一切都很好，但优化并没有收敛(或为此做任何事情(。当我调用lsq_res.x时，我可以看到参数与最初的猜测相同。我怀疑最小化函数本身的问题(def-fun(x，t，y(：…(并且不确定我是否正确地传递了变量。非常感谢您的帮助，因为这也允许我将其应用于其他型号！

import numpy as np
import matplotlib.pyplot as plt
import math
from scipy.optimize import least_squares
def Intensity(x_data, A21, T21, A22, T22, A23, T23, y0):
I_model=A21*np.exp(-x_data/T21)+A22*np.exp(-x_data/T22)+A23*np.exp(-x_data/T23)+y0
return I_model

#generate example data set (should be replaced by load of csv data file)
def gen_data(t, b1, c1, b2, c2, b3, c3, y0, noise=0, n_outliers=0, random_state=0):
y = b1 * np.exp(-t / c1) + b2 * np.exp(-t / c2) + b3 * np.exp(-t / c3)+y0
rnd = np.random.RandomState(random_state)
error = noise * rnd.randn(t.size)
outliers = rnd.randint(0, t.size, n_outliers)
error[outliers] *= 10
return y + error
# these are the parameters used to calculate the function, correspond to my first guess
y0 = 0.5
b1 = 0.25
c1 = .01
b2 = 0.4
c2 = .3
b3 = 0.35
c3 = 10
t_min = -3
t_max = 2
n_points = 1000
x_data = np.logspace(t_min, t_max, n_points)
y_data = gen_data(x_data, b1, c1, b2, c2, b3, c3, y0, noise=0.1, n_outliers=3) 
# the following is the minimization function where the appropriate model needs to be entered in the return line. 
def fun(x, t, y):
return Intensity(x_data, A21, T21, A22, T22, A23, T23, y0) - y_data 
x0 = np.array([A21, T21, A22, T22, A23, T23, y0]) # give starting values for the fit parameters in the model

res_lsq = least_squares(fun, x0, args=(x_data, y_data)) #this performs the actual minimization of
y_lsq = gen_data(x_data, *res_lsq.x)

Answer 1

当我看到这个问题时，您试图为Intensity函数找到最佳参数。

我建议您使用scipy的curve_fit函数。

import numpy as np
from scipy.optimize import curve_fit
def Intensity(x_data, A21, T21, A22, T22, A23, T23, y0 ):
I_model=A21*np.exp(-x_data/T21)+A22*np.exp(-x_data/T22)+A23*np.exp(-x_data/T23)+y0
return I_model

#generate example data set (should be replaced by load of csv data file)
def gen_data(t, b1, c1, b2, c2, b3, c3, y0, noise=0, n_outliers=0, random_state=0):
y = b1 * np.exp(-t / c1) + b2 * np.exp(-t / c2) + b3 * np.exp(-t / c3)+y0
rnd = np.random.RandomState(random_state)
error = noise * rnd.randn(t.size)
outliers = rnd.randint(0, t.size, n_outliers)
error[outliers] *= 10
return y + error
#%%
# these are the parameters used to calculate the function, correspond to my first guess
y0 = 0.5
b1 = 0.25
c1 = .01
b2 = 0.4
c2 = .3
b3 = 0.35
c3 = 10
t_min = -3
t_max = 2
n_points = 1000
x_data = np.logspace(t_min, t_max, n_points)
y_data = gen_data(x_data, b1, c1, b2, c2, b3, c3, y0, noise=0.1, n_outliers=3)
res = curve_fit(Intensity, x_data, y_data) 

文档可以在这里找到。我希望我没有误解你的问题。

如果您想最小化函数，您应该提供静态参数值和变量的初始猜测。在您的问题中，没有给出参数值。