我想使用lmfit将ellipsometric数据拟合到复杂模型中。两个测量参数psi
和delta
是复杂函数rho
中的变量。
我可以尝试使用shared parameters或picewise方法将问题分离到实部和虚部,但是有没有办法直接用复杂函数来做? 仅适合函数的实部工作,但是当我定义复杂的残差函数时,我得到:
TypeError:没有为复数定义排序关系。
下面是我的实际功能拟合代码以及我解决复杂拟合问题的尝试:
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 我解决了虚部和实部的分离变量的问题。数据形状应为[[imaginary_data],[real_data]],目标函数必须返回1D数组。
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 ...
答案 0 :(得分:1)
lmfit FAQ建议只使用numpy.ndarray.view
同时获取实部和虚部,这意味着您无需手动分离实部和虚部。
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()