The zoo package includes the "yearmon" class to facilitate such manipulations. Here are a few solutions assuming you store you series in a zoo variable:
# test data library(zoo) z <- zoo(1001:1100, as.Date(101:200))[-(45:55)] # Solution 1. tapply produces indexes of last of month tt <- time(z) z[ c(tapply(seq_along(tt), as.yearmon(tt), tail, 1)) ] # If we want to create a last variable which corresponds # to last in sas then do it this slightly longer way: # Solution 2 tt <- time(z) last <- seq_along(tt) %in% tapply(seq_along(tt), as.yearmon(tt), tail, 1) z[last] # Solution 3. another solution with a last variable. f(x) is # vector same length as x with all 0's except last element is 1. tt <- time(z) f <- function(x) replace(0*x, length(x), 1) last <- ave(seq_along(tt), as.yearmon(tt), FUN = f) z[last] In all these solutions the last point in the series is always included. We have not assumed that every day is necessarily included in your series but if every day is included then even simpler solutions are possible. On 8/29/07, Shubha Vishwanath Karanth <[EMAIL PROTECTED]> wrote: > Hi R users, > > > > Is there a function in R, which does some calculation only for the month > end in a daily data?... In other words, is there a command in R, > equivalent to "last." function in SAS? > > > > BR, Shubha > > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help@stat.math.ethz.ch 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@stat.math.ethz.ch 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.
