On Thu, 21 Sep 2006, Thorsten Muehge wrote: > > Hello R Experts, > I want to aggregate parameters by week. But our production week ends Friday > night instead of Sunday Night which is the default value in R.
The default in ISO8601, not just in R, but that is %W, not %U as used below. > In order to solve the problem I want to substract two days from the current > data and than use the R function > > test$week<-format(test$dates,"%U"); > > with a test&dates format equal to "2006-09-21". > > How do I substract the two days from the test$dates column in the > data.frame? You have not told us what class test$dates is! Assuming it is "Date", test$dates-2. *However*, to do what you ask, you need to add 1: > dates <- seq(as.Date("2006-09-21"), by=1, len=7) > format(dates+1, "%U") [1] "38" "38" "39" "39" "39" "39" "39" There is a potential problem here at year ends (there is anyway in the ISO8601 definition). Another way is just (unclass(dates) - 2) %/% 7 which orders weeks across years. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UK Fax: +44 1865 272595 ______________________________________________ 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.