r - R %dopar% 比 %do% 慢

Question

我正在尝试对 %dopar% 进行一个小测试，但结果总是比较慢。这是一些虚拟代码和输出。我在win7x64 core i7-2600k上运行这个...

library(foreach)
library(doSNOW)
registerDoSNOW(makeCluster(3, type = "SOCK"))

N <- 3*(10^4)

system.time(foreach(i = 1:N) %do% {sum(rnorm(N))})
system.time(foreach(i = 1:N) %dopar% {sum(rnorm(N))} )

这是输出：

> system.time(foreach(i = 1:N) %do% {sum(rnorm(N))})
 user  system elapsed 
90.39    0.00   90.42 

> system.time(foreach(i = 1:N) %dopar% {sum(rnorm(N))} )
user  system elapsed 
17.00    0.89  177.11