比较线性，非线性和不同参数化非线性模型的方法

Question

我搜索了一种比较线性，非线性和不同参数化非线性模型的方法。为了这：

#Packages
library(nls2)
library(minpack.lm)

# Data set - Diameter in function of Feature and Age
Feature<-sort(rep(c("A","B"),22))
Age<-c(60,72,88,96,27,
36,48,60,72,88,96,27,36,48,60,72,
88,96,27,36,48,60,27,27,36,48,60,
72,88,96,27,36,48,60,72,88,96,27,
36,48,60,72,88,96)
Diameter<-c(13.9,16.2,
19.1,19.3,4.7,6.7,9.6,11.2,13.1,15.3,
15.4,5.4,7,9.9,11.7,13.4,16.1,16.2,
5.9,8.3,12.3,14.5,2.3,5.2,6.2,8.6,9.3,
11.3,15.1,15.5,5,7,7.9,8.4,10.5,14,14,
4.1,4.9,6,6.7,7.7,8,8.2)
d<-dados <- data.frame(Feature,Age,Diameter)
str(d)

我将创建三个不同的模型，两个具有特定参数化的非线性模型和一个线性模型。在我的例子中，假设每种模式的所有系数都是显着的（并且不考虑实际结果）。

# Model 1 non-linear
e1<- Diameter ~ a1 * Age^a2 
#Algoritm Levenberg-Marquardt
m1 <-  nlsLM(e1, data = d,
     start = list(a1 = 0.1, a2 = 10),
     control = nls.control(maxiter = 1000))

# Model 2 linear
m2<-lm(Diameter ~ Age, data=d)

# Model 3 another non-linear
e2<- Diameter ~ a1^(-Age/a2)
m3 <-  nls2(e2, data = d, alg = "brute-force",
     start = data.frame(a1 = c(-1, 1), a2 = c(-1, 1)),
     control = nls.control(maxiter = 1000))

现在，我的想法是比较“更好”的模型，尽管每个模型的性质不同，比我尝试比例测量，为此我使用每个模型的每个均方误差比较数据集中的总平方误差，当一个使这个我有比较模型1和2：

## MSE approach (like pseudo R2 approach)

#Model 1
SQEm1<-summary(m1)$sigma^2*summary(m1)$df[2]# mean square error of model 
SQTm1<-var(d$Diameter)*(length(d$Diameter)-1)#total square error in data se
R1<-1-SQEm1/SQTm1
R1

#Model 2
SQEm2<-summary(m2)$sigma^2*summary(m2)$df[2]# mean square error of model 
R2<-1-SQEm2/SQTm1
R2

在我的弱观点中，模型1比模型2“更好”。我的问题是，这种方法听起来是否正确？有没有办法比较这些模型类型？

提前致谢！

Answer 1

#First cross-validation approach ------------------------------------------

#Cross-validation model 1
set.seed(123) # for reproducibility

n <- nrow(d)
frac <- 0.8
ix <- sample(n, frac * n) # indexes of in sample rows

e1<- Diameter ~ a1 * Age^a2 
#Algoritm Levenberg-Marquardt
m1 <-  nlsLM(e1, data = d,
     start = list(a1 = 0.1, a2 = 10),
     control = nls.control(maxiter = 1000), subset = ix)# in sample model

BOD.out <- d[-ix, ] # out of sample data
pred <- predict(m1, new = BOD.out)
act <- BOD.out$Diameter
RSS1 <- sum( (pred - act)^2 )
RSS1
#[1] 56435894734

#Cross-validation model 2
m2<-lm(Diameter ~ Age, data=d,, subset = ix)# in sample model
BOD.out2 <- d[-ix, ] # out of sample data
pred <- predict(m2, new = BOD.out2)
act <- BOD.out2$Diameter
RSS2 <- sum( (pred - act)^2 )
RSS2
#[1] 19.11031

# Sum of squares approach -----------------------------------------------
deviance(m1)
#[1] 238314429037

deviance(m2)
#[1] 257.8223

基于gfgm和G.Grothendieck的评论，RSS2具有较低的误差，RSS1和比较偏差（m2）和偏差（m2）也比模型2好于模型1。