Displaying 1 result from an estimated 1 matches for "dipr".
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dipl
2002 Oct 02
0
Alternative to slow double for() loop
...peed things up.
The program looks like this:
for (j in 1:m) {
for (i in 1:n) {
times<-comp.list[[j]][which(comp.list[[j]]$V1==i),]
T<-ncol(times)
Y<-times$V2
Y<-data.matrix(Y)
cova<-subset(times, select=V3:V16)
cova<-data.matrix(cova)
pr<-exp(cova%*%beta)/(1+exp(cova%*%beta))
dipr<-diag(c(pr[1,1], pr[2,1]))
dipr1<-dipr-(pr%*%t(pr))
A<-diag(c(dipr1[1,1],dipr1[2,2]))
D<-t(cova)%*%A
V<-A
u1<-D%*%solve(V)%*%(Y-pr)
u<-u+u1
usq<-usq+(u1%*%t(u1))
dvd<-dvd+D%*%solve(V)%*%t(D)
u.list[[j]]<-u
usq.list[[j]]<-usq
dvd.list[[j]]<-dvd
}}
where j are a n...