优化拟合系数以获得更好的拟合

Question

我是运行使用 minpack.lm 程序包的非线性最小二乘法。

但是，对于数据中的每个组，我想优化（最小化）拟合参数，类似于 Python 的 minimize 函数。

The minimize() function is a wrapper around Minimizer for running an optimization problem. It takes an objective function (the function that calculates the array to be minimized), a Parameters object, and several optional arguments.

之所以需要这个，是因为我想根据得到的拟合参数优化拟合函数，找到数据中能同时拟合两个groups的全局拟合参数。

这是我目前在 group 中的拟合方法ps,

df <- data.frame(y=c(replicate(2,c(rnorm(10,0.18,0.01), rnorm(10,0.17,0.01))), 
                                c(replicate(2,c(rnorm(10,0.27,0.01), rnorm(10,0.26,0.01))))),
                         DVD=c(replicate(4,c(rnorm(10,60,2),rnorm(10,80,2)))),
                         gr = rep(seq(1,2),each=40),logic=rep(c(1,0),each=40))

这组ps的拟合方程为

fitt <- function(data) {
  fit <- nlsLM(y~pi*label2*(DVD/2+U1)^2,
               data=data,start=c(label2=1,U1=4),trace=T,control = nls.lm.control(maxiter=130))
}

library(minpack.lm)
library(plyr)  # will help to fit in groups

fit <- dlply(df, c('gr'), .fun = fitt)  #,"Die" only grouped by Waferr

> fit
$`1`
Nonlinear regression model
  model: y ~ pi * label2 * (DVD/2 + U1)^2
   data: data
   label2        U1 
2.005e-05 1.630e+03 
$`2`
label2      U1 
 2.654 -35.104

我想知道是否有任何函数可以优化平方和以获得最适合两个组的函数ps。我们可能会说您已经拥有最佳拟合参数作为残差平方和，但我知道 minimizer 可以做到这一点，但我还没有找到我们可以在 R 中做到这一点的任何类似示例。

ps。我编了数字和拟合线。

Answer 1

不确定 r，但具有共享参数的最小二乘法通常很容易实现。

一个简单的 python 示例如下：

import matplotlib
matplotlib.use('Qt4Agg')
from matplotlib import pyplot as plt

from random import random
from scipy import optimize
import numpy as np

#just for my normal distributed errord
def boxmuller(x0,sigma):
    u1=random()
    u2=random()
    ll=np.sqrt(-2*np.log(u1))
    z0=ll*np.cos(2*np.pi*u2)
    z1=ll*np.cos(2*np.pi*u2)
    return sigma*z0+x0, sigma*z1+x0

#some non-linear function
def f0(x,a,b,c,s=0.05):
    return a*np.sqrt(x**2+b**2)-np.log(c**2+x)+boxmuller(0,s)[0]

# residual function for least squares takes two data sets. 
# not necessarily same length
# two of three parameters are common
def residuals(parameters,l1,l2,dataPoints):
    a,b,c1,c2 = parameters
    set1=dataPoints[:l1]
    set2=dataPoints[-l2:]
    distance1 = [(a*np.sqrt(x**2+b**2)-np.log(c1**2+x))-y for x,y in set1]
    distance2 = [(a*np.sqrt(x**2+b**2)-np.log(c2**2+x))-y for x,y in set2]
    res = distance1+distance2
    return res

xList0=np.linspace(0,8,50)
#some xy data
xList1=np.linspace(0,7,25)
data1=np.array([f0(x,1.2,2.3,.33) for x in xList1])
#more xy data using different third parameter
xList2=np.linspace(0.1,7.5,28)
data2=np.array([f0(x,1.2,2.3,.77) for x in xList2])
alldata=np.array(zip(xList1,data1)+zip(xList2,data2))

# rough estimates
estimate = [1, 1, 1, .1]
#fitting; providing second length is actually redundant
bestFitValues, ier= optimize.leastsq(residuals, estimate,args=(len(data1),len(data2),alldata))
print bestFitValues


fig = plt.figure()
ax = fig.add_subplot(111)
ax.scatter(xList1, data1)
ax.scatter(xList2, data2)
ax.plot(xList0,[f0(x,bestFitValues[0],bestFitValues[1],bestFitValues[2] ,s=0) for x in xList0])
ax.plot(xList0,[f0(x,bestFitValues[0],bestFitValues[1],bestFitValues[3] ,s=0) for x in xList0])


plt.show()

#output
>> [ 1.19841984  2.31591587  0.34936418  0.7998094 ]

如果需要，您甚至可以自己进行最小化。如果您的参数 space 表现良好，即近似抛物线最小值，那么简单的 Nelder Mead method 就可以了。

优化拟合系数以获得更好的拟合

Optimized fitting coefficients for better fitting

r

curve-fitting

nls

nonlinear-optimization