ask for help
On 07 Aug 2014, at 11:16 , jim holtman <jholtman at gmail.com> wrote:
rle
...with a little tinkering, like
m <- c(1,cumsum(rle(a)$lengths)+1) m
[1] 1 4 6 12 16 18 23 34 then look at every 2nd element, discarding the last.
Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd.mes at cbs.dk Priv: PDalgd at gmail.com