[R] sort

Jim Lemon drj|m|emon @end|ng |rom gm@||@com
Fri May 15 05:58:14 CEST 2020


Hi Val,
Your problem is keeping the orders straight. You want dates decreasing
and names increasing:

DF1<-read.table(text="name ddate
  A  2019-10-28
  A  2018-01-25
  A  2020-01-12
  A  2017-10-20
  B  2020-11-20
  B  2019-10-20
  B  2017-05-20
  B  2020-01-20
  C  2009-10-01  ",header=TRUE)
DF1$Time<-as.POSIXct(DF1$ddate , format = "%Y-%m-%d")
# get dates in decreasing order
DF2<-DF1[order(DF1$Time,decreasing=TRUE),]
# create the final output with the names in increasing order
DF3<-data.frame(name=sort(unique(DF2$name)))
# create a function that returns the first diff of a vector
getdate1diff<-function(x)
 return(ifelse(length(x)>1,abs(diff(x))[1],0))
# apply the function by the unique names in DF2
DF3$date1diff<-by(DF2$Time,DF2$name,getdate1diff)
DF3

Jim

On Fri, May 15, 2020 at 1:00 PM Val <valkremk using gmail.com> wrote:
>
> HI All,
> I have a sample of data frame
> DF1<-read.table(text="name ddate
>   A  2019-10-28
>   A  2018-01-25
>   A  2020-01-12
>   A  2017-10-20
>   B  2020-11-20
>   B  2019-10-20
>   B  2017-05-20
>   B  2020-01-20
>   c  2009-10-01  ",header=TRUE)
>
> 1. I want sort by name and ddate on decreasing order and the output
> should like as follow
>    A  2020-01-12
>    A  2019-01-12
>    A  2018-01-25
>    A  2017-10-20
>    B  2020-11-21
>   B  2020-11-01
>   B  2019-10-20
>   B  2017-05-20
>   c  2009-10-01
>
> 2.  Take the top two rows by group( names) and the out put should like
>    A  2020-01-12
>    A  2019-01-12
>    B  2020-11-21
>    B  2020-11-01
>     c  2009-10-01
>
> 3.  Within each group (name) get the date difference  between the
> first and second rows dates. If a group has only one row then the
> difference should be 0
>
> The final out put is
> Name diff
>    A  365
>     B  20
>     C  0
>
> Here is my attempt and have an issue at the sorting
> DF1$DTime <- as.POSIXct(DF1$ddate , format = "%Y-%m-%d")
> DF2 <- DF1[order(DF1$name, ((as.Date(DF1$DTime, decreasing = TRUE)))), ]
>
> not working
> Any help?
>
> Thank you
>
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