Question

I have the example data and model

x<-rep(seq(0,100,by=1),10)
y<-15+2*rnorm(1010,10,4)*x+rnorm(1010,20,100)
id<-NULL
for (i in 1:10){
  id<-c(id, rep(i,101))}
dtfr<-data.frame(x=x,y=y, id=id)
library(nlme)
with (dtfr, summary(lme((y)~x,random=~1+x|id, na.action=na.omit  )))
model.mx<-with (dtfr, (lme((y)~x,random=~1+x|id, na.action=na.omit  )))
pd<-predict(model.mx, newdata=data.frame(x=0:100),level=0)
with (dtfr, plot(x, y))
lines(0:100,predict(model.mx, newdata=data.frame(x=0:100),level=0), col="darkred", lwd=7)

How can I extract the modelled intercept and slope of each individual ID and plot the individual trajectories of each ID?

Était-ce utile?

La solution

Not sure what you want to do because all your coefficients are almost identical:

> coef(model.mx)
   (Intercept)        x
1     54.88302 19.18001
2     54.88298 19.18000
3     54.88299 19.18000
4     54.88299 19.18000
5     54.88302 19.18001
6     54.88300 19.18000
7     54.88301 19.18000
8     54.88300 19.18000
9     54.88299 19.18000
10    54.88300 19.18000

Maybe your real data gives you more different results. If it's the case, I would use abline inside a mapply call:

with (dtfr, plot(x, y))
mapply(abline,a=coef(model.mx)[,1],b=coef(model.mx)[,2], col=1:10)

Here's the result. Since all coeffcients are almost the same, the lines are plotted on top of each other. You only see the last one. enter image description here

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