Just so. I got until 'split' but was stuck on how to get the breaks ... Thank you!
Joh jim holtman wrote: > Is this what you want: > >> testVector <- c(12,32,NA,NA,56,NA,78,65,87,NA,NA,NA,90) >> # get the breaks at the NAs >> xb <- cumsum(!is.na(testVector)) >> split(seq(length(testVector)), xb) > $`1` > [1] 1 > > $`2` > [1] 2 3 4 > > $`3` > [1] 5 6 > > $`4` > [1] 7 > > $`5` > [1] 8 > > $`6` > [1] 9 10 11 12 > > $`7` > [1] 13 > > > On Wed, Oct 28, 2009 at 7:57 AM, Johannes Graumann > <johannes_graum...@web.de> wrote: >> Dear all, >> >> Is there an efficient way to get this list >>> testList <- list(c(1),c(2,3,4),c(5,6),c(7),c(8),c(9,10,11,12),c(13)) >> >> from this vector >>> testVector <- c(12,32,NA,NA,56,NA,78,65,87,NA,NA,NA,90) >> ? >> >> Basically the vector should be grouped, such that non-NA and all >> following NAs end up in one group. >> >> Thanks for any hint, >> >> Joh >> >> ______________________________________________ >> 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.