[R] Using lapply in R data table
Ista Zahn
istazahn at gmail.com
Mon Sep 26 19:05:51 CEST 2016
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_rod at 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]]
>
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