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. > -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? ______________________________________________ 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.