Performance with foreach, doSNOW, and snowfall
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Is it just me, or does the performance of the foreach package with a doSNOW backend operating on a socket grid suck?Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
Here at work, I am helping to setup a cluster of Windows machines for distributed R processing. We have lots of researchers running code that takes hours to complete and are essentially large for loops with lots of analysis in between. These guys and gals are not hard core programmers, so there is lots of interest in foreach (as opposed to something like RMPI).
I have successfully setup a POC grid between mutliple machines using sockets and public key authentication. Assuming we use this, I’ll post a how-to, as there is not much on the web on how to get it working on Windows.
In the meantime, I am testing performance. There is something going on with foreach that I do not understand. Performance numbers are really bad.
Can anyone explain what is going on here?
> require(doSNOW)EDIT: HA! Figured it out. foreach is not very efficient in communicating tasks as compared to par*apply(). The time to communicate the process overwhelmed the actual processing time.
Loading required package: doSNOW
Loading required package: foreach
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Loading required package: iterators
Loading required package: snow
> require(snowfall)
Loading required package: snowfall
>
> sfInit(parallel=TRUE,socketHosts=rep(“localhost”,3))
R Version: R version 2.15.0 (2012-03-30)
snowfall 1.84 initialized (using snow 0.3-9): parallel execution on 3 CPUs.
> cl = sfGetCluster()
>
> f = function(x) {
+ sum = 0
+ for (i in seq(1,x)) sum = sum + i
+ return(sum)
+ }
>
> registerDoSNOW(cl)
>
> out = vector(“logical”,length=10000)
> system.time( (for (i in seq(1,10000)) out[i]=f(i) ))
user system elapsed
25.99 0.00 25.99
>
> system.time( (out = lapply(seq(1,10000),f) ))
user system elapsed
26.55 0.00 26.55
>
> system.time( (out = parLapply(cl,seq(1,10000),f) ))
user system elapsed
0.02 0.00 15.85
>
> system.time( (out = foreach(i=seq(1,10000)) %dopar% f(i) ))
user system elapsed
6.64 0.42 98.31
>
> getDoParWorkers()
[1] 3
When I change the code to this, it runs fast (about the same as parLapply()):
> system.time( (out = foreach(i=seq(0,9),.combine=’c’) %dopar% {
+ apply(as.array(seq(i*1000+1,(i+1)*1000)),1,f)
+ }))
user system elapsed
0.00 0.00 14.03
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