CODEWITHSUNDEEP
rforeachparallel-processingplyr
S. Oued
S. Oued

Reputation: 13

%dopar% in R does not work properly

I just start to use the foreach and %dopar% methodes for parallel processing in R , but the results I'm getting are confusing and not the same as a for loop; here is the code I used to test those methodes and resultes I'm getting: 

library(plyr); library(doParallel); library(foreach)

cs <- makeCluster(2)
registerDoParallel(cs)

sfor_start <- Sys.time()
s_for=as.numeric()
for (i in 1:1000) {
  s_for[i] = sqrt(i)
}
print(Sys.time() - sfor_start)

sdopar_start <- Sys.time()
sdopar=as.numeric()
foreach(k=1:1000) %dopar% {
  sdopar[k] = sqrt(k)
}
print(Sys.time() - sdopar_start)

And here the results:

> s_for[1:10]; sdopar[1:10]
 [1] 1.000000 1.414214 1.732051 2.000000 2.236068 2.449490 2.645751 2.828427 3.000000 3.162278
 [1] NA NA NA NA NA NA NA NA NA NA

Thanks in advance :)

Upvotes: 0

Views: 9352

Answers (1)

F. Priv&#233;
F. Priv&#233;

Reputation: 11738

Please read the documentation of functions before saying that they don't work.

foreach works more like a lapply than a for-loop.

So, for example, foreach(k=1:1000) %dopar% sqrt(k) gives the same result as lapply(1:1000, sqrt).

Yet, it is true that you can modify global variable when using foreach SEQUENTIALLY. Yet, when using parallelism, the vector sdopar is copied to each "cluster" so that you modify a copy, not the initial object.

So, you'll have to do as mentioned by @ChiPak with option .combine = c or using do.call(sdopar, c) afterwards.

PS: Always initialize the vector you fill iteratively (for efficiency of not growing a vector), for example like this: s_for <- double(1000).

Upvotes: 8

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