You can remove missing values with: zm <- aggregate(cambio, as.yearmon, mean, na.rm = TRUE)
Its not clear what your second question is asking. If you want the series to have a Date class rather than yearmon class with the 1st of the month then: zd <- zm time(zd) <- as.Date(time(zm)) or zd <- aggregate(zm, as.Date, force) On Wed, Apr 15, 2009 at 5:28 PM, manta <mantin...@libero.it> wrote: > > Ok, using > > mcambio <- aggregate(cambio, as.yearmon, mean) > > works perfectly!! Should I worry about the missing values or not anyway? And > then I go to the following question. From monthly data to daily using a > specific formula? > -- > View this message in context: > http://www.nabble.com/From-daily-series-to-monthly-and-viceversa-tp23064454p23067500.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. > ______________________________________________ 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.