[R] seq() problem with chron

Jeff Newmiller jdnewm|| @end|ng |rom dcn@d@v|@@c@@u@
Thu Sep 6 09:54:25 CEST 2018


See FAQ 7.31... chron uses floating point representation, and there is 
some error accumulating. I also think there may be at least one bug in 
chron::seq.dates(), but I think POSIXct is significantly better than chron 
anywway so I don't intend to debug chron.

#############################
# with chron, avoiding repeated addition of fractional days
library(chron)
dt1 <- chron("02/20/13", "00:00:00")
dt2 <- chron("07/03/18", "15:30:00")
n <- round( 24*4*as.numeric( dt2 - dt1 ) )
length(dt)
#> [1] 1
dt <- dt1 + seq( 0, n ) / 24 / 4
dt[length(dt)]
#> [1] (07/03/18 15:30:00)
# with POSIXct
Sys.setenv( TZ = "GMT" )
DT1 <- as.POSIXct( "02/20/2013 00:00:00", format = "%m/%d/%Y %H:%M:%S" )
DT2 <- as.POSIXct( "07/03/2018 15:30:00", format = "%m/%d/%Y %H:%M:%S" )
# POSIXct is represented as seconds, so no fractions are used
DT <- seq( DT1, DT2, by = as.difftime( 15, units="mins" ) )
DT[length(DT)]
#> [1] "2018-07-03 15:30:00 GMT"

#' Created on 2018-09-06 by the [reprex package](http://reprex.tidyverse.org) (v0.2.0).
#############################


On Thu, 6 Sep 2018, Waichler, Scott R wrote:

> Hi,
>
> I encountered the problem below where the last value in the chron vector created with seq() should have a time of 15:30, but instead has 15:15.  What causes this and how can I make sure that the last value in the chron vector is the same as the "to" value in seq()?
>
> library(chron)
> dt1 <- chron("02/20/13", "00:00:00")
> dt2 <- chron("07/03/18", "15:30:00")
> dt <- seq(from=dt1, to=dt2, by=1/(24*4))
> dt[length(dt)]
> #[1] (07/03/18 15:15:00)
>
> Thanks,
> Scott Waichler
> Pacific Northwest National Laboratory
> scott.waichler using pnnl.gov
>
> ______________________________________________
> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
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>

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