[R] Using lapply in R data table

[Bert Gunter]

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 <istazahn at 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_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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>> and provide commented, minimal, self-contained, reproducible code.
>
> ______________________________________________
> R-help at 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.