[R] Using lapply in R data table
Bert Gunter
bgunter.4567 at gmail.com
Mon Sep 26 22:27:05 CEST 2016
Ista:
Aha -- now I see the point. My bad. You are right. I was careless.
However, cut() with ifelse() might simplify the code a bit and/or make
it more readable. To be clear, this is just a matter of taste; e.g.
using your data and a data frame instead of a data table:
> DT <- within(DT,
exposure <- {
f <-cut(fini,as.Date(c("2000-01-01","2006-01-01","2006-06-30","2006-12-21")),
labels= letters[1:3])
ifelse(f == "a", 1,
ifelse( f == "c", .5,
difftime(as.Date("2007-01-01"), fini, units="days")/365.25))
}
)
> DT
id fini group exposure f
1 2 2005-04-20 A 1.0000000 a
2 2 2005-04-20 A 1.0000000 a
3 2 2005-04-20 A 1.0000000 a
4 5 2006-02-19 B 0.8651608 b
5 5 2006-06-29 B 0.5092402 b
6 7 2006-10-08 A 0.5000000 c
7 7 2006-10-08 A 0.5000000 c
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 12:07 PM, Ista Zahn <istazahn at gmail.com> wrote:
> On Mon, Sep 26, 2016 at 2:48 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote:
>> I thought that that was a typo from the OP, as it disagrees with his
>> example. But the labels are arbitrary, so in fact cut() will do it
>> whichever way he meant.
>
> I don't see how cut will do it, at least not conveniently. Consider
> this slightly altered example:
>
> library(data.table)
> DT <- data.table(
> id = rep(c(2, 5, 7), c(3, 2, 2)),
> fini = rep(as.Date(c('2005-04-20',
> '2006-02-19',
> '2006-06-29',
> '2006-10-08')),
> c(3, 1, 1, 2)),
> group = rep(c("A", "B", "A"), c(3, 2, 2)) )
>
> 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]
>
> DT
>
> ## id fini group exposure
> ## 1: 2 2005-04-20 A 1.0000000
> ## 2: 2 2005-04-20 A 1.0000000
> ## 3: 2 2005-04-20 A 1.0000000
> ## 4: 5 2006-02-19 B 0.8651608
> ## 5: 5 2006-06-29 B 0.5092402
> ## 6: 7 2006-10-08 A 0.5000000
> ## 7: 7 2006-10-08 A 0.5000000
>
> Best,
> Ista
>
>>
>> -- 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 11:37 AM, Ista Zahn <istazahn at gmail.com> wrote:
>>> On Mon, Sep 26, 2016 at 1:59 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote:
>>>> This seems like a job for cut() .
>>>
>>> I thought that at first two, but the middle group shouldn't be .87 but rather
>>>
>>> exposure" = "2007-01-01" - "fini"
>>>
>>> so, I think cut alone won't do it.
>>>
>>> Best,
>>> Ista
>>>>
>>>> (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.
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