I'm not able to replicate your problem. Here's what I get. See if this is what you want:
> dt$date<-strptime(as.character(dt$datetime),"%d%b%Y") > dt datetime date 1 01OCT1987:00:00:00.000 1987-10-01 2 12APR2004:00:00:00.000 2004-04-12 3 01DEC1987:00:00:00.000 1987-12-01 4 01OCT1975:00:00:00.000 1975-10-01 5 01AUG1979:00:00:00.000 1979-08-01 6 26JUN2003:00:00:00.000 2003-06-26 7 01JAN1900:00:00:00.000 1900-01-01 8 13MAY1998:00:00:00.000 1998-05-13 9 30SEP1998:00:00:00.000 1998-09-30 cheers, -Girish =========================================================== premmad wrote: > > It works but what i need is the result also as a column . > I tried using the following code . > dt$new<-strptime(as.character(dt$datetime),"%d%b%Y. > It shows the following error > > Error in `$<-.data.frame`(`*tmp*`, "Sa_dt", value = list(sec = c(0, 0, : > replacement has 9 rows, data has 14. > > Please help me to solve this > -- View this message in context: http://www.nabble.com/Datetime-conversion-tp25503138p25506493.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.