使用Python和lmfit拟合复杂的模型? [英] Fitting complex model using Python and lmfit?

问题描述

我想使用LMFit使椭偏数据适合复杂模型.测得的两个参数psi和delta是复杂函数rho中的变量.

I would like to fit ellipsometric data to complex model using LMFit. Two measured parameters, psi and delta, are variables in a complex function rho.

我可以尝试使用逐项方法，但是有有什么办法可以直接使用复杂的功能吗? 仅拟合函数的实部效果很好，但是当我定义复杂的残差函数时，我得到:

I could try with separating problem to real and imaginary part with shared parameters or piecewise approach, but is there any way to do it directly with complex function? Fitting only real part of function works beautifully, but when I define complex residual function I get:

TypeError:没有为复数定义排序关系.

TypeError: no ordering relation is defined for complex numbers.

下面是我的实函数拟合代码，也是我尝试解决复杂拟合问题的代码:

Below is my code for real function fitting and my attempt at tackling complex fit problem:

    from __future__ import division
    from __future__ import print_function
    import numpy as np
    from pylab import *
    from lmfit import minimize, Parameters, Parameter, report_errors


    #=================================================================
    #             MODEL

    def r01_p(eps2, th):
        c=cos(th)
        s=(sin(th))**2

        stev= sqrt(eps2) * c - sqrt(1-(s / eps2))
        imen= sqrt(eps2) * c + sqrt(1-(s / eps2))
        return stev/imen

    def r01_s(eps2, th):
        c=cos(th)
        s=(sin(th))**2

        stev= c - sqrt(eps2) * sqrt(1-(s/eps2))
        imen= c + sqrt(eps2) * sqrt(1-(s/eps2))
        return stev/imen


    def rho(eps2, th):
        return r01_p(eps2, th)/r01_s(eps2, th)

    def psi(eps2, th):
        x1=abs(r01_p(eps2, th))
        x2=abs(r01_s(eps2, th))
        return np.arctan2(x1,x2)

    #=================================================================
    #                   REAL FIT
    #

    #%%

    # generate data from model  
    th=linspace(deg2rad(45),deg2rad(70),70-45)
    error=0.01
    var_re=np.random.normal(size=len(th), scale=error)
    data = psi(2,th) + var_re

    # residual function
    def residuals(params, th, data):
        eps2 = params['eps2'].value

        diff = psi(eps2, th) - data
        return diff

    # create a set of Parameters
    params = Parameters()
    params.add('eps2',   value= 1.0,  min=1.5, max=3.0)


    # do fit, here with leastsq model
    result = minimize(residuals, params, args=(th, data),method="leastsq")

    # calculate final result
    final = data + result.residual

    # write error report
    report_errors(params)


    # try to plot results
    th, data, final=rad2deg([th, data, final])
    try:
        import pylab
        clf()
        fig=plot(th, data, 'r o',
                 th, final, 'b')
        setp(fig,lw=2.)
        xlabel(r'$\theta$ $(^{\circ})$', size=20)
        ylabel(r'$\psi$ $(^{\circ})$',size=20)

        show()

    except:
        pass

    #%%
    #=================================================================
    #                   COMPLEX FIT

    # TypeError: no ordering relation is defined for complex numbers

    """
    # data from model with added noise   
    th=linspace(deg2rad(45),deg2rad(70),70-45)
    error=0.001
    var_re=np.random.normal(size=len(th), scale=error)
    var_im=np.random.normal(size=len(th), scale=error) * 1j

    data = rho(4-1j,th) + var_re + var_im


    # residual function
    def residuals(params, th, data):
        eps2 = params['eps2'].value

        diff = rho(eps2, th) - data
        return np.abs(diff)

    # create a set of Parameters
    params = Parameters()
    params.add('eps2',   value= 1.5+1j,  min=1+1j, max=3+3j)


    # do fit, here with leastsq model
    result = minimize(residuals, params, args=(th, data),method="leastsq")

    # calculate final result
    final = data + result.residual

    # write error report
    report_errors(params)
    """
    #=================================================================

修改: 我用虚部和实部分开的变量解决了问题.数据的形状应为[[imaginary_data]，[real_data]]，目标函数必须返回一维数组.

I solved problem with separated variables for imaginary and real part. Data should be shaped as [[imaginary_data],[real_data]], objective function must return 1D array.

def objective(params, th_data, data):
    eps_re  = params['eps_re'].value
    eps_im  = params['eps_im'].value
    d       = params['d'].value

    residual_delta = data[0,:] - delta(eps_re - eps_im*1j, d, frac, lambd, th_data)
    residual_psi   = data[1,:] - psi(eps_re - eps_im*1j, d, frac, lambd, th_data)

    return np.append(residual_delta,residual_psi)

# create a set of Parameters
params = Parameters()
params.add('eps_re',   value= 1.5,  min=1.0,      max=5  )
params.add('eps_im',   value= 1.0,  min=0.0,      max=5  )
params.add('d',        value= 10.0,   min=5.0,    max=100.0   )


# All available methods
methods=['leastsq','nelder','lbfgsb','anneal','powell','cobyla','slsqp']
# Chosen method
#metoda='leastsq'

# run the global fit to all the data sets
result = minimize(objective, params, args=(th_data,data),method=metoda))

....

return ...

推荐答案

The lmfit FAQ suggests simply taking both real and imaginary parts by using numpy.ndarray.view, which means you don't need to go through the separation of the real and imaginary parts manually.

def residuals(params, th, data):
    eps2 = params['eps2'].value

    diff = rho(eps2, th) - data
    # The only change required is to use view instead of abs.
    return diff.view()

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