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
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--
==============================================
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
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