Question

假设我有一个函数f()和一个向量d

f <- function(x) dexp(x, 2)
d <- runif(10, 1, 5)

现在我想执行一个for循环

dnew <- numeric(length(d))
for (i in seq_along(dnew)){
   dnew[i] <- f(d[i])
}

。如何并行执行此操作？

Answer 1

没有for循环的示例代码更快：

dnew2 <- f(d)          # 'f()' and 'd' from question
all.equal(dnew, dnew2) # 'dnew' from question 
[1] TRUE

library(microbenchmark)
microbenchmark('for loop' = for (i in seq_along(dnew)){ dnew[i] <- f(d[i]) },
               'vectorized' = { dnew2 = f(d) })
Unit: microseconds
       expr    min      lq     mean  median     uq    max neval
   for loop 15.639 16.4455 17.66640 17.0045 18.089 43.938   100
 vectorized  1.249  1.3140  1.44039  1.3845  1.516  2.424   100

它可以与 foreach 并行化：

library(foreach)
library(doParallel); registerDoParallel(2)
dnew3 <- foreach(i=seq_along(dnew), .combine=c) %dopar% {
    f(d[i])
}
all.equal(dnew, dnew3)
[1] TRUE

并行化版本较慢，因为并行开销大于收益。

microbenchmark('for loop' = for (i in seq_along(dnew)){ dnew[i] <- f(d[i]) },
               'foreach' = { dnew3 <- foreach(i=seq_along(dnew), .combine=c) %dopar% {
                                      f(d[i]) } 
                            })
Unit: microseconds
     expr       min        lq        mean     median         uq       max neval
 for loop    17.799    22.048    31.01027    32.7615    37.0945    67.265   100
  foreach 11875.845 13003.558 13576.64759 13427.1015 14041.3455 17782.638   100

如果f()需要花费更多时间进行评估，则 foreach 版本会更快：

f <- function(x){
    Sys.sleep(.3)
    dexp(x, 2)
}
microbenchmark('for loop' = for (i in seq_along(dnew)){ dnew[i] <- f(d[i]) },
               'foreach' = {dnew3 <- foreach(i=seq_along(dnew), .combine=c) %dopar% {
                                     f(d[i]) }
                }, times=2)
Unit: seconds
     expr      min       lq     mean   median       uq      max neval
 for loop 3.004271 3.004271 3.004554 3.004554 3.004837 3.004837     2
  foreach 1.515458 1.515458 1.515602 1.515602 1.515746 1.515746     2

Answer 2

简单的循环

a <- function(x) {dexp(x,2)}
d<- runif(10,1,5)
d
dnew < - numeric(length(d))
for (i in 1: length(dnew)){
dnew[i]<- a(d[i])
}
dnew

并行版本

library(doParallel)
dnew < - numeric(length(d))
no_cores <- detectCores() - 1  
registerDoParallel(cores=no_cores)  
cl <- makeCluster(no_cores, type="FORK")  
dnew <- clusterApply(cl=cl, x=d, fun = a)  
stopCluster(cl) 
dnew

看看这个博客的帖子：https://www.r-bloggers.com/lets-be-faster-and-more-parallel-in-r-with-doparallel-package/

希望有帮助！

我如何在R中并行运行for循环

2 个答案: