Try this: > system.time(Sys.sleep(60)) user system elapsed 0.00 0.00 60.05 > pt <- proc.time(); Sys.sleep(60); proc.time() - pt user system elapsed 0.00 0.00 60.01
On Sat, Feb 27, 2010 at 9:33 PM, Ravi Varadhan <rvarad...@jhmi.edu> wrote: > > Hi, > > The `system.time(expr)' command provide 3 different times for evaluating the > expression `expr'; the first two are user and system CPUs and the third one > is total elapsed time. Suppose I want to compare two different computational > procedures for performing the same task, which component of `system.time' is > most meaningful in the sense that it most accurately reflects the > computational effort of the algorithm, and does not depend upon the > idiosyncrasies of the operating system. > > I have always been using the first component of `system.time', which is the > user CPU. Should I use the sum of user and system CPU or is the total > elapsed time a better measure? I would appreciate UseR's feedback on this. > > Thanks very much. > > Best, > Ravi. > ____________________________________________________________________ > > Ravi Varadhan, Ph.D. > Assistant Professor, > Division of Geriatric Medicine and Gerontology > School of Medicine > Johns Hopkins University > > Ph. (410) 502-2619 > email: rvarad...@jhmi.edu > > ______________________________________________ > 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.
