Thank you! Your interpretation was correct. I'll remember to be more specific next time. Thank you for being patient and understanding.
Jacques On 5/21/08, jim holtman <[EMAIL PROTECTED]> wrote: > Not sure exactly what you want since you did not provide any data or an > example of the expected out' Here is my interpretation of what you were > asking: > >> x <- sample(1:20) >> x > [1] 6 16 8 1 17 11 2 19 18 5 15 13 3 20 9 14 7 10 12 4 >> >> cumsum(x) / seq_along(x) > [1] 6.000000 11.000000 10.000000 7.750000 9.600000 9.833333 8.714286 > 10.000000 10.888889 10.300000 > [11] 10.727273 10.916667 10.307692 11.000000 10.866667 11.062500 10.823529 > 10.777778 10.842105 10.500000 >> > > > On Wed, May 21, 2008 at 9:48 PM, Jacques Wagnor <[EMAIL PROTECTED]> > wrote: > >> Dear List, >> >> Does there exist a function that calculates a cumulative average? >> Neither running() from library(gregmisc) nor running.mean() from >> library(igraph) seems to be able to give a cumulative average. >> >> Any help or pointers would be greatly appreciated. >> >> Jacques >> >> ______________________________________________ >> 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<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 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.
