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Approach for comparing linear, non-linear and different parameterization non-linear models


Leave one out cross validation with lm function in RSave multiple linear regression modelsIs there a way to mimic the nls nls.control(warnOnly=TRUE) when using nlsLM?Using generalized linear models to compare group means in RCross-validation for non-linear regression using nls in RHow to calculate mean values from a linear model in R?Modeling a repeated measures logistic growth curveNon-linear regression vs log modelUse VAE approach in a Linear Regression model to compute output uncertaintyExtract and add to the data values of the probability density function based on a stan linear model






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








0















I search for one approach for comparing linear, non-linear and different parameterization non-linear models. For this:



#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)


I will create three different models, two non-linear models with specific parametization and one linear model. In my example
a suppose that all the coefficients of each mode were significant (and not considering real results).



# 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))


Now, my idea is comparing the "better" model despite the different nature of each model, than I try a proportional measure
and for this I use each mean square error of each model comparing of total square error in data set, when a make this I have if
a comparing model 1 and 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


In my weak opinion model 1 is "better" that model 2. My question is, does this approach sounds correct? Is there any way to compare these models types?



Thanks in advance!










share|improve this question





















  • 2





    this way of comparing models doesn't penalize models for complexity and risks overfitting the data. you would be safer comparing your models via cross-validation

    – gfgm
    Mar 28 at 14:01






  • 1





    IF the models have the same number of parameters as is the case here then you can just use the sum of squares of residuals: deviance(m1); deviance(m2) where lower is better. Also graph the fit superimposed on the data and that may make it obvious which model fits best. Be sure to sort the data on Age so that the plots come out right.

    – G. Grothendieck
    Mar 28 at 17:29


















0















I search for one approach for comparing linear, non-linear and different parameterization non-linear models. For this:



#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)


I will create three different models, two non-linear models with specific parametization and one linear model. In my example
a suppose that all the coefficients of each mode were significant (and not considering real results).



# 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))


Now, my idea is comparing the "better" model despite the different nature of each model, than I try a proportional measure
and for this I use each mean square error of each model comparing of total square error in data set, when a make this I have if
a comparing model 1 and 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


In my weak opinion model 1 is "better" that model 2. My question is, does this approach sounds correct? Is there any way to compare these models types?



Thanks in advance!










share|improve this question





















  • 2





    this way of comparing models doesn't penalize models for complexity and risks overfitting the data. you would be safer comparing your models via cross-validation

    – gfgm
    Mar 28 at 14:01






  • 1





    IF the models have the same number of parameters as is the case here then you can just use the sum of squares of residuals: deviance(m1); deviance(m2) where lower is better. Also graph the fit superimposed on the data and that may make it obvious which model fits best. Be sure to sort the data on Age so that the plots come out right.

    – G. Grothendieck
    Mar 28 at 17:29














0












0








0


1






I search for one approach for comparing linear, non-linear and different parameterization non-linear models. For this:



#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)


I will create three different models, two non-linear models with specific parametization and one linear model. In my example
a suppose that all the coefficients of each mode were significant (and not considering real results).



# 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))


Now, my idea is comparing the "better" model despite the different nature of each model, than I try a proportional measure
and for this I use each mean square error of each model comparing of total square error in data set, when a make this I have if
a comparing model 1 and 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


In my weak opinion model 1 is "better" that model 2. My question is, does this approach sounds correct? Is there any way to compare these models types?



Thanks in advance!










share|improve this question
















I search for one approach for comparing linear, non-linear and different parameterization non-linear models. For this:



#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)


I will create three different models, two non-linear models with specific parametization and one linear model. In my example
a suppose that all the coefficients of each mode were significant (and not considering real results).



# 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))


Now, my idea is comparing the "better" model despite the different nature of each model, than I try a proportional measure
and for this I use each mean square error of each model comparing of total square error in data set, when a make this I have if
a comparing model 1 and 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


In my weak opinion model 1 is "better" that model 2. My question is, does this approach sounds correct? Is there any way to compare these models types?



Thanks in advance!







r linear-regression lm nls non-linear-regression






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 28 at 23:01









Steve

4,8101 gold badge8 silver badges31 bronze badges




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asked Mar 28 at 13:44









LeprechaultLeprechault

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  • 2





    this way of comparing models doesn't penalize models for complexity and risks overfitting the data. you would be safer comparing your models via cross-validation

    – gfgm
    Mar 28 at 14:01






  • 1





    IF the models have the same number of parameters as is the case here then you can just use the sum of squares of residuals: deviance(m1); deviance(m2) where lower is better. Also graph the fit superimposed on the data and that may make it obvious which model fits best. Be sure to sort the data on Age so that the plots come out right.

    – G. Grothendieck
    Mar 28 at 17:29













  • 2





    this way of comparing models doesn't penalize models for complexity and risks overfitting the data. you would be safer comparing your models via cross-validation

    – gfgm
    Mar 28 at 14:01






  • 1





    IF the models have the same number of parameters as is the case here then you can just use the sum of squares of residuals: deviance(m1); deviance(m2) where lower is better. Also graph the fit superimposed on the data and that may make it obvious which model fits best. Be sure to sort the data on Age so that the plots come out right.

    – G. Grothendieck
    Mar 28 at 17:29








2




2





this way of comparing models doesn't penalize models for complexity and risks overfitting the data. you would be safer comparing your models via cross-validation

– gfgm
Mar 28 at 14:01





this way of comparing models doesn't penalize models for complexity and risks overfitting the data. you would be safer comparing your models via cross-validation

– gfgm
Mar 28 at 14:01




1




1





IF the models have the same number of parameters as is the case here then you can just use the sum of squares of residuals: deviance(m1); deviance(m2) where lower is better. Also graph the fit superimposed on the data and that may make it obvious which model fits best. Be sure to sort the data on Age so that the plots come out right.

– G. Grothendieck
Mar 28 at 17:29






IF the models have the same number of parameters as is the case here then you can just use the sum of squares of residuals: deviance(m1); deviance(m2) where lower is better. Also graph the fit superimposed on the data and that may make it obvious which model fits best. Be sure to sort the data on Age so that the plots come out right.

– G. Grothendieck
Mar 28 at 17:29













1 Answer
1






active

oldest

votes


















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


Based in gfgm and G. Grothendieck comments, RSS2 has lower error that RSS1 and comparing deviance(m2) and deviance(m2) too, than model 2 is better than model 1.






share|improve this answer


























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    #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


    Based in gfgm and G. Grothendieck comments, RSS2 has lower error that RSS1 and comparing deviance(m2) and deviance(m2) too, than model 2 is better than model 1.






    share|improve this 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


      Based in gfgm and G. Grothendieck comments, RSS2 has lower error that RSS1 and comparing deviance(m2) and deviance(m2) too, than model 2 is better than model 1.






      share|improve this answer





























        1














        1










        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


        Based in gfgm and G. Grothendieck comments, RSS2 has lower error that RSS1 and comparing deviance(m2) and deviance(m2) too, than model 2 is better than model 1.






        share|improve this answer















        #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


        Based in gfgm and G. Grothendieck comments, RSS2 has lower error that RSS1 and comparing deviance(m2) and deviance(m2) too, than model 2 is better than model 1.







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        edited Mar 28 at 19:07

























        answered Mar 28 at 19:00









        LeprechaultLeprechault

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