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? [[alternative HTML version deleted]] ______________________________________________ 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.
