Well, one way to do it is via regex's -- no splitting and recombining needed. Note: This will convert a factor into a character vector.
> z <- c("11/7/2016", "14-07-16") > z <- gsub("-([[:digit:]]{2})-([[:digit:]]{2})", "/\\1/20\\2",z) ## /\ is / and \ > z [1] "11/7/2016" "14/07/2016" I leave it to you as an exercise to either convert 7 to 07 or vice-versa if you want to do this. Note, if you have spaces sprinkled inconsistently around your separators, you'll have to work a bit harder with your regex. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Sat, Nov 2, 2019 at 7:25 PM <reichm...@sbcglobal.net> wrote: > R-Help Forum > > > > I have a data set that contains a date field but the dates are in two > formats > > > > 11/7/2016 dd/mm/yyyy > > 14-07-16 dd-mm-yy > > > > How would I go about correcting this problem. Should I separate the dates, > format them , and then recombine? > > > > Sincerely > > > > Jeff Reichman > > (314) 457-1966 > > > > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > 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. > [[alternative HTML version deleted]] ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.