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 <rvaradhan at 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: rvaradhan at jhmi.edu
>
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