用python中的数值模型进行曲线拟合 [英] curve fitting with a numerical model in python

问题描述

我有一些适合模型的数据点.我的模型不是定义为方程，而是定义为3个方程的数值解.我的模型定义如下:

I have some data points to fit with a model. My model is not defined as an equation but as a numerical solution of 3 equations. My model is defined as below:

def eq(q):
    z1=q[0]
    z2=q[1]
    H=q[2]

    F=empty((3))
    F[0] = ((J*(1-(D*(1-(1-8*a*T/D**3)**(1/3)))**(b)/Lx))*sin(z1-z2))+(H*sin(z1-pi/4))+(((3.6*10**5)/2)*sin(2*z1))
    F[1] = ((-J*(1-(D*(1-(1-8*a*T/D**3)**(1/3)))**(b)/Lx))*sin(z1-z2))+(H*sin(z2-pi/4))+(((3.6*10**5)/2)*sin(2*z2))
    F[2] = cos(z1-pi/4)
    return F

guess1=array([2.35,0.2,125000])
z=fsolve(eq,guess1)
Hc=z[2]*(1-(T/Tb)**(1/2))

那个

D=10**(-8)
a=2.2*10**(-28)
Lx=4.28*10**(-9)

和 J ， b ， Tb 是参数，而 z1 ， z2 ， H 是变量

and J, b, Tb are parameters and z1, z2, H are variables

我的数据点是:

T=[10, 60, 110, 160, 210, 260, 300]

Hc=[0.58933, 0.5783, 0.57938, 0.58884, 0.60588, 0.62788, 0.6474]

如何根据具有数据点的拟合模型找到 J ， b ， Tb ?

how can I find J, b, Tb according to fitting model with data points?

推荐答案

您可以使用 scipy.optimize.curve_fit .您需要了解，curve_fit只关心函数的输入和输出，因此您需要使用以下方式进行定义:

You can use scipy.optimize.curve_fit. You need to understand that curve_fit will only care about the input and the output of your function, such that you need to define it this way :

def func(x, *params):
    ....
    return y

然后，您可以应用 curve_fit (

Then you can apply curve_fit (https://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.curve_fit.html):

popt, pcov = curve_fit(func, x, y_data)

我写信并举例说明您的情况.有100种方法可以更有效地编写它.但是，我试图弄清楚这一点，以便您可以理解您所给与我所提议之间的联系.

I wrote and example taking your case. There is 100 ways to write it more efficiently. However, I tried to make it clear so you can understand the link between what you gave and what I propose.

import numpy as np
from scipy.optimize import fsolve, curve_fit

def eq(q, T, J, b): 

    z1 = q[0]
    z2 = q[1]
    H = q[2]

    D=10**(-8)
    a=2.2*10**(-28)
    Lx=4.28*10**(-9)    


    F = np.empty((3))
    F[0] = ((J*(1-(D*(1-(1-8*a*T/D**3)**(1/3)))**(b)/Lx))*np.sin(z1-z2))+(H*np.sin(z1-np.pi/4))+(((3.6*10**5)/2)*np.sin(2*z1))
    F[1] = ((-J*(1-(D*(1-(1-8*a*T/D**3)**(1/3)))**(b)/Lx))*np.sin(z1-z2))+(H*np.sin(z2-np.pi/4))+(((3.6*10**5)/2)*np.sin(2*z2))
    F[2] = np.cos(z1-np.pi/4)
    return F

def func(T,J,b,Tb):
    Hc = []
    guess1 = np.array([2.35,0.2,125000])
    for t in T :
        z = fsolve(eq, guess1, args = (t, J, b))
        Hc.append(z[2]*(1-(t/Tb)**(1/2)))
    return Hc

T = [10, 60, 110, 160, 210, 260, 300]
Hc_exp = [0.58933, 0.5783, 0.57938, 0.58884, 0.60588, 0.62788, 0.6474]

p0 = (1,1,100)
popt, pcov = curve_fit(func, T, Hc_exp, p0)

J = popt[0]
b = popt[1]
Tb = popt[2]

似乎很难做到合适.为了改善这一点，您可以通过 p0 向参数添加初始猜测，或为参数添加界限(请参见 curve_fit 文档).收敛后，您可能会在估计的参数上出现错误.

It seems that the fit is hard to be done. To improve that, you can add an initial guess to the parameters through p0, or adding bounds to the parameters (see curve_fit documentation). Once it converges, you can have the error on the estimated parameters.

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