Displaying 1 result from an estimated 1 matches for "haz2surv".
2007 Jan 19
1
Error in basehaz function ?
...<- 1 - surv[i] / surv[i-1]
Using this rule I achieve satisfiable results with the two following
functions :
surv2haz <- function(surv) {
haz <- surv
haz[1] <- 1 - surv[1]
for(i in c(2:length(surv)))
{
haz[i] <- 1 - surv[i] / surv[i - 1]
}
return(haz)
}
haz2surv <- function(haz) {
surv <- haz
surv[1] <- 1 - haz[1]
for(i in c(2:length(haz)))
{
surv[i] <- (1 - haz[i]) * surv[i-1]
}
return(surv)
}
If I'm right, wouldn't it be a good idea to change the basehaz
function, to avoid misleading the overconfident user...