Thanks.
I also did some parallel computing. Here are some time I spent by using
'snowfall' package on intel centrino 2 laptop (2 cores) on windows 7.
user system elapsed
0.240.05 2442.77
Libo
On Wed, Feb 22, 2012 at 12:53 AM, Martin Maechler <
maech...@stat.math.ethz.ch> wrote:
> >>
On 22/02/2012 07:53, Martin Maechler wrote:
"LS" == Libo Sun
on Tue, 21 Feb 2012 21:09:54 -0700 writes:
> Thanks. Shall I sum the user time and system time, which
> roughly equals to the elapsed time?
No. Rather just use the user time,
i.e. proc.time()[[1]]
system.time
> "LS" == Libo Sun
> on Tue, 21 Feb 2012 21:09:54 -0700 writes:
> Thanks. Shall I sum the user time and system time, which
> roughly equals to the elapsed time?
No. Rather just use the user time,
i.e. proc.time()[[1]]
system.time()[[1]] etc
That's typically good
Thanks. Shall I sum the user time and system time, which roughly equals to
the elapsed time?
I also tried to improve the code by using 'cmpfun(myfunction)' in
'compiler' package, however, it doesn't help too much.
Libo
On Tue, Feb 21, 2012 at 8:11 PM, R. Michael Weylandt <
michael.weyla...@gmail
The time relationships aren't strictly linear between any of the three
measures, but *very generally* I've interpreted them as something
like:
User: stuff you do (i.e., doing all the commands)
System: stuff at the OS level (memory allocations and whatnot)
Elapsed: Clock time
None is a great measu
Hi all,
I got this time for my code,
> proc.time()-pt
user systemelapsed
132541.743 0.004 132533.526
As you can see, there is huge difference btw elapsed time and system time.
Does that mean lots of I/O? Or some bad coding?
Thanks,
Libo
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