Hi Tammy, I suspect that the first 4 digits compose the year, so with Year_Month<-c(2010*100+9:11,2011*100+1:2) #check composition Year_Month
ym<-(Year_Month%/%100*12+Year_Month%%100) ym-ym[1] should give the difference in months. Differences in Days are a bit trickier, since not all months have an equal number of days. Something like xd<-as.Date(paste(Year_Month%/%100,Year_Month%%100,"01",sep="-")) xd-xd[1] deals with that. Cheers Am 01.06.2012 13:31, schrieb Tammy Ma: > > HI, R-Users: > > I got a questions. have been struggling so long time.... > > I have this data: > >> m1$Year_Month > 201009 201010 201011 201101 201102 >> min(m1$Year_Month) > 201009 > > I want to calculate the following two answers, how do I program it? > >> difference in Month????? > [1] 0 1 2 4 5 > >> difference in Days????? > 0 31 61 .... > > Thank you in advance!!! > > Tammy > > [[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. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus ______________________________________________ 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.