[R] seq() problem with chron

Daniel Nordlund djnord|und @end|ng |rom gm@||@com
Thu Sep 6 09:07:02 CEST 2018

On 9/5/2018 10:00 PM, 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
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This is not a chron problem, it is a floating-point arithmetic problem 
(basically, FAQ 7.31).  You are adding incrementing by 1/96, which can't 
be represent exactly in binary representation.  So, when you expected 
that you should get a time of 15:30, it is slightly larger and the 
sequence is stopped at 15:15.

You could change dt2 to be chron("07/03/18", "15:31:00").  Or or you 
could use POSIX datetimes with something like the following, where the 
increment 900 is the number of seconds in 15 minutes.

dt1 <- strptime("02/20/13 00:00:00", "%m/%d/%y %H:%M:%S")
dt2 <- strptime("07/03/18 15:30:00", "%m/%d/%y %H:%M:%S")
dt <- seq(from=dt1, to=dt2, by=900)

There might also be some useful functions in the lubridate package.

Hope this is helpful,


Daniel Nordlund
Port Townsend, WA  USA

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