library(xts) period.apply() might also be helpful.
Michael On Tue, Nov 6, 2012 at 7:32 PM, Rui Barradas <ruipbarra...@sapo.pt> wrote: > Hello, > > You should provide us with a data example. Since you haven't, look at the > following code and see if you understand it. > > # Make up some data > x <- Sys.time() + (1:1000)*15 > y <- rnorm(1000) > > brks <- cut(x, breaks = "hour") # hour breaks > tapply(y, brks, mean) # hourly means of 'y' > > > Hope this helps, > > Rui Barradas > Em 06-11-2012 18:57, B. Bahar escreveu: > >> Hello, >> >> I have much more than one milion tempreture is gained each 15 seconds. So >> each 15 seconds, I have one data. >> How could I calculate avarage datas in each hour in R? >> >> If you could help me. >> >> [[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. > > > ______________________________________________ > 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.
