[R] Running segmented on grouped data and collating model parameters in a data frame

[Vito M. R. Muggeo]

dear Sumi,
I am not familiar with the dplyr package (%>%..), however if you want to 
fit the model for each subject times freq interaction, a simple for loop 
will suffice.
Here possible code:

Assuming d is the dataframe, something like

subj<-levels(d$subject)
fr<-unique(d$freq)

#new dataframe where the estimates will be stored
newd<-expand.grid(subj=subj, freq=fr)
newd<-cbind(newd, matrix(-99,nrow(newd),9))
names(newd)[-(1:2)]<-c(paste0("slope",1:3), 
paste0(c("CI.inf.","CI.sup."), rep(paste0("slope",1:3),each=2)))

library(segmented)

#fit the model for each subject x freq combination
for(i in subj){
       for(j in fr){
             current.d<-subset(d, subject==i & freq==j)
             if(nrow(current.d)>0){
               o<-lm(Ldp_p~L2, data=current.d)
               os<-try(segmented(o, ~L2, psi=c(20,50)), silent=TRUE)
               if(class(os)[1]!="try-error"){
                      est.slope<-as.vector(slope(os)[[1]][,1]) #point est
                      ci.slope<-as.vector(t(slope(os)[[1]][,4:5])) #CI
                      newd[newd$subj==i & newd$freq==j, 
-c(1:2)]<-c(est.slope, ci.slope)
                                        }
                      }

       }
}


However it seems that you can be interested in segmented *mixed* 
models.. I mean, rather than estimating a segmented model for each 
subject, you could estimate a single model assuming random effects (for 
each model parameter, including the breakpoint) for each subject:

Have a look to

http://journals.sagepub.com/doi/abs/10.1177/1471082X13504721

Let me know (not on the R list) if you are interested in relevant code

best,
vito


Il 03/06/2018 18:03, Sumitrajit Dhar ha scritto:
> Hi Folks,
> 
> I am trying to teach myself how to solve the problem described below but am
> running out of time. Hence the plea for help. Thanks in advance.
> 
> Here is my data frame.
> 
>> t
> # A tibble: 12 x 12
>     subject  ageGrp ear   hearingGrp sex    freq    L2   Ldp Phidp    NF
> SNR   Ldp_p
>     <fct>    <fct>  <fct> <fct>      <fct> <int> <int> <dbl> <dbl> <dbl>
> <dbl>   <dbl>
>   1 HALAF573 A      L     A          F         2     0 -19.6 197.  -28.5
> 8.88  2.10e-6
>   2 HALAF573 A      L     A          F         2     2 -18.7 203.  -22.0
> 3.25  2.31e-6
>   3 HALAF573 A      L     A          F         2     4 -29.1 255.  -27.4
>   -1.64  7.04e-7
>   4 HALAF573 A      L     A          F         2     6 -12.4 174.  -12.2
>   -0.206 4.78e-6
>   5 HALAF573 A      L     A          F         4     0 -28.6 232.  -26.7
>   -1.87  7.45e-7
>   6 HALAF573 A      L     A          F         4     2 -27.2 351.  -28.8
> 1.59  8.68e-7
>   7 HALAF573 A      L     A          F         4     4 -20.4  26.2 -35.0
>   14.6   1.92e-6
>   8 HALAF573 A      L     A          F         4     6 -20.0  85.1 -29.8
> 9.75  2.00e-6
>   9 HALAF573 A      L     A          F         8     0 -22.8  39.2 -22.1
>   -0.689 1.45e-6
> 10 HALAF573 A      L     A          F         8     2 -14.5  13.4 -20.7
> 6.23  3.76e-6
> 11 HALAF573 A      L     A          F         8     4 -17.3 345.  -21.6
> 4.30  2.73e-6
> 12 HALAF573 A      L     A          F         8     6 -14.1 320.  -21.7
> 7.59  3.94e-6
> 
> # Note that there are more levels of L2 (31 in total)  and 344 other
> subjects but I truncated the frame for posting.
> 
> # I want to do this:  t %>%  group_by(freq) %>% [run segmented] %>% [create
> a data frame with [subject, freq, breakpoint1, breakpoint2, slope1, slope2,
> slope3, L2 when Ldp_p == 0].
> 
> # Also note that ultimately I will be grouping by "subject, freq".
> 
> # I can run the models and get believable results. The following run on a
> data frame with L2 between 0 and 60.
> 
> out.lm <- lm(Ldp_p ~ L2, data = t)
> o <- segmented(out.lm, seg.Z = ~L2, psi = list(L2 = c(20,45)),
>                 control = seg.control(display = FALSE)
> 
> o$psi
> #        Initial     Est.    St.Err
> #psi1.L2      20 30.78256 0.5085192
> #psi2.L2      45 53.16390 0.4671701
> 
> slope(o)
> slope(o)
> $L2
> #              Est.    St.Err. t value   CI(95%).l   CI(95%).u
> #slope1  1.2060e-06 1.6606e-07  7.2622  8.6397e-07  1.5480e-06
> #slope2  1.0708e-05 2.9196e-07 36.6770  1.0107e-05  1.1309e-05
> #slope3 -4.5791e-06 1.3694e-06 -3.3439 -7.3995e-06 -1.7588e-06
> 
> Thanks again for bailing me out.
> 
> Regards,
> Sumit
> 
> 	[[alternative HTML version deleted]]
> 
> ______________________________________________
> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
> 

-- 
==============================================
Vito M.R. Muggeo
Dip.to Sc Econom, Az e Statistiche
Università di Palermo
viale delle Scienze, edificio 13
90128 Palermo - ITALY
tel: 091 23895240
fax: 091 485726
http://dssm.unipa.it/vmuggeo
Associate Editor, Statistical Modelling
Chair, Statistical Modelling Society