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Frank S.

2016-Sep-26 15:28 UTC

[R] Using lapply in R data table

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]]

Ista Zahn

2016-Sep-26 17:05 UTC

head link

[R] Using lapply in R data table

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]]
>
> ______________________________________________
> 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.

Bert Gunter

2016-Sep-26 17:59 UTC

head link

[R] Using lapply in R data table

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]]
>>
>> ______________________________________________
>> 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.
>
> ______________________________________________
> 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.

R help - Sep 2016 - Using lapply in R data table

[R] Using lapply in R data table

[R] Using lapply in R data table

[R] Using lapply in R data table