[R] as.Data with minutes resolution

[Gabor Grothendieck]

You can use chron:

   library(chron)
   time1 <- c("03/08/08-11:00","03/08/08-11:10")
   diff(as.chron(time1, "%d/%m/%y-%H:%M"))

Also you can use POSIXct but in that case be careful
about time zones.  See R News 4/1.

On Mon, Nov 10, 2008 at 7:38 AM, Ronaldo Reis-Jr. <chrysopa at gmail.com> wrote:
> Hi,
>
> I have a vetor os dates with day and hour:minutes.
>
>> time1 <- c("03/08/08-11:00","03/08/08-11:10")
>> time1 <- as.Date(time1,"%d/%m/%y-%R")
>> summary(time1)
>        Min.      1st Qu.       Median         Mean      3rd Qu.
> Max.
> "2008-08-03" "2008-08-03" "2008-08-03" "2008-08-03" "2008-08-03"
> "2008-08-03"
>> time1[2]-time1[1]
> Time difference of 0 days
>
> I need to get the difference between these minutes, my expected result in
> this operation is 10 minutes.
>
> How I make this in R?
>
> Thanks
> Ronaldo
>
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