This seems like a job for cut() . (I made DT a data frame to avoid loading the data table package. But I assume it would work with a data table too, Check this, though!)
> DT <- within(DT, exposure <- > cut(fini,as.Date(c("2000-01-01","2006-01-01","2006-06-30","2006-12-21")), > labels= c(1,.87,.5))) > DT id fini group exposure 1 2 2005-04-20 A 1 2 2 2005-04-20 A 1 3 2 2005-04-20 A 1 4 5 2006-02-19 B 0.87 5 5 2006-02-19 B 0.87 6 7 2006-10-08 A 0.5 7 7 2006-10-08 A 0.5 (but note that exposure is a factor, not numeric) Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Mon, Sep 26, 2016 at 10:05 AM, Ista Zahn <istaz...@gmail.com> wrote: > Hi Frank, > > lapply(DT) iterates over each column. That doesn't seem to be what you want. > > There are probably better ways, but here is one approach. > > DT[, exposure := vector(mode = "numeric", length = .N)] > DT[fini < as.Date("2006-01-01"), exposure := 1] > DT[fini >= as.Date("2006-01-01") & fini <= as.Date("2006-06-30"), > exposure := difftime(as.Date("2007-01-01"), fini, units="days")/365.25] > DT[fini >= as.Date("2006-07-01"), exposure := 0.5] > > Best, > Ista > > On Mon, Sep 26, 2016 at 11:28 AM, Frank S. <f_j_...@hotmail.com> wrote: >> Dear all, >> >> I have a R data table like this: >> >> DT <- data.table( >> id = rep(c(2, 5, 7), c(3, 2, 2)), >> fini = rep(as.Date(c('2005-04-20', '2006-02-19', '2006-10-08')), c(3, 2, >> 2)), >> group = rep(c("A", "B", "A"), c(3, 2, 2)) ) >> >> >> I want to construct a new variable "exposure" defined as follows: >> >> 1) If "fini" earlier than 2006-01-01 --> "exposure" = 1 >> 2) If "fini" in [2006-01-01, 2006-06-30] --> "exposure" = "2007-01-01" - >> "fini" >> 3) If "fini" in [2006-07-01, 2006-12-31] --> "exposure" = 0.5 >> >> >> So the desired output would be the following data table: >> >> id fini exposure group >> 1: 2 2005-04-20 1.00 A >> 2: 2 2005-04-20 1.00 A >> 3: 2 2005-04-20 1.00 A >> 4: 5 2006-02-19 0.87 B >> 5: 5 2006-02-19 0.87 B >> 6: 7 2006-10-08 0.50 A >> 7: 7 2006-10-08 0.50 A >> >> >> I have tried: >> >> DT <- DT[ , list(id, fini, exposure = 0, group)] >> DT.new <- lapply(DT, function(exposure){ >> exposure[fini < as.Date("2006-01-01")] <- 1 # 1st case >> exposure[fini >= as.Date("2006-01-01") & fini <= >> as.Date("2006-06-30")] <- difftime(as.Date("2007-01-01"), fini, >> units="days")/365.25 # 2nd case >> exposure[fini >= as.Date("2006-07-01") & fini <= as.Date("2006-12-31")] >> <- 0.5 # 3rd case >> exposure # return value >> }) >> >> >> But I get an error message. >> >> Thanks for any help!! >> >> >> Frank S. >> >> >> [[alternative HTML version deleted]] >> >> ______________________________________________ >> R-help@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. > > ______________________________________________ > R-help@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. ______________________________________________ R-help@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.