Try this: sapply(0:(length(DateTime3)-1), function(i)as.character(strptime(start, "%m/%d/%Y %H:%M") + i * interval))
On Wed, Oct 13, 2010 at 8:51 AM, dpender <d.pen...@civil.gla.ac.uk> wrote: > > I am trying to convert an array from numeric values back to date and time > format. The code I have used is as follows; > > for (i in 0:(length(DateTime3)-1)) { > DateTime3[i] <- (strptime(start, "%m/%d/%Y %H:%M")+ i*interval) > > where start <- [1] "1/1/1981 00:00" > > However the created array (DateTime3) contains [1,] 347156400 347157600 > 347158800 347160000 347161200 347162400 347163600 NA > > Does anyone know how I can change DateTime 3 to the same format as start? > > Thanks, > > Doug > > -- > View this message in context: > http://r.789695.n4.nabble.com/Date-Time-Objects-tp2993524p2993524.html > Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > R-help@r-project.org mailing list > 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. > -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40" S 49° 16' 22" O [[alternative HTML version deleted]]
______________________________________________ R-help@r-project.org mailing list 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.