On Tue, 3 Mar 2009, Romain Francois wrote:
Prof Brian Ripley wrote:
The caching is in the disc system: you need to find and read the package
metadata for every package. AFAIK it is not easy to flush the disc cache,
but quite easy to overwrite it with later reads. (Google for more info.)
Thanks for the info, I'll try to find my way with these directions.
If you are not concerned about validity of the installed packages you could
skip the tests and hence the reads.
Your times are quite a bit slower than mine, so a faster disc system might
help. Since my server has just been rebooted (for a new kernel), with all
of CRAN and most of BioC I get
system.time( packs <- .packages( all = T ) )
user system elapsed
0.518 0.262 25.042
system.time( packs <- .packages( all = T ) )
user system elapsed
0.442 0.080 0.522
length(packs)
[1] 2096
There's a similar issue when installing packages: the Perl code reads the
indices from every visible package to resolve links, and that can be slow
the first time.
On Tue, 3 Mar 2009, Romain Francois wrote:
Hello,
The first time in a session I call .packages( all.available = T ), it
takes a long time (I have many packages installed from CRAN):
system.time( packs <- .packages( all = T ) )
user system elapsed
0.738 0.276 43.787
When I call it again, the time is now much reduced, so there must be some
caching somewhere. I would like to try to reduce the time it takes the
first time, but I have not been able to identify where the caching takes
place, and so how I can remove it to try to improve the running time
without the caching. Without this, I have to restart my computer each time
to vanish the caching to test a new version of the function (this is not
going to happen)
Here is the .packages function, I am suspicious about this part : "ans <-
c(ans, nam)" which grows the ans vector each time a suitable package is
found, this does not sound right.
It's OK as there are only going to be ca 2000 packages. Try profiling
this: .readRDS and grepl take most of the time.
I usually do not trust the result of the profiler when a for loop is
involved, as it tends to miss the point (or maybe I am).
Here are the data for the actual example (repeated for this message):
Rprof()
system.time( packs <- .packages( all = T ) )
user system elapsed
0.447 0.078 0.525
Rprof(NULL)
summaryRprof()
$by.self
self.time self.pct total.time total.pct
"grepl" 0.18 34.6 0.18 34.6
".readRDS" 0.12 23.1 0.20 38.5
".packages" 0.08 15.4 0.50 96.2
"close.connection" 0.04 7.7 0.04 7.7
"close" 0.02 3.8 0.06 11.5
"file.exists" 0.02 3.8 0.02 3.8
"gc" 0.02 3.8 0.02 3.8
"gzfile" 0.02 3.8 0.02 3.8
"list" 0.02 3.8 0.02 3.8
"system.time" 0.00 0.0 0.52 100.0
"file.path" 0.00 0.0 0.02 3.8
$by.total
total.time total.pct self.time self.pct
"system.time" 0.52 100.0 0.00 0.0
".packages" 0.50 96.2 0.08 15.4
".readRDS" 0.20 38.5 0.12 23.1
"grepl" 0.18 34.6 0.18 34.6
"close" 0.06 11.5 0.02 3.8
"close.connection" 0.04 7.7 0.04 7.7
"file.exists" 0.02 3.8 0.02 3.8
"gc" 0.02 3.8 0.02 3.8
"gzfile" 0.02 3.8 0.02 3.8
"list" 0.02 3.8 0.02 3.8
"file.path" 0.02 3.8 0.00 0.0
$sampling.time
[1] 0.52
there is little tiime unaccounted for, and 0.38 sec is going in
.readRDS and grepl. Whereas
system.time({
ans <- character(0)
for(i in 1:2096) ans <- c(ans, "foo")
})
takes 0.024 secs, negligible here (one profiler tick).
Consider this script below,
Whether profiling works in other examples is beside the point here.
[...]
--
Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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