Question

我想将一些并行的R代码放入函数中，但是当在函数内部而不是函数外部时，该代码的运行速度要慢得多。

该代码采用一个坐标矩阵并将其转换为sf对象。下面的演示代码将相同的过程运行两次。

简单代码
相同的代码，但功能范围内

示例代码：

# Packages
library(parallel)
library(sf)
library(pbapply)

# Sample Data
mat = matrix(runif(1e6,max = 1e5), ncol = 2)
seqs = 1:(nrow(mat)-1)
ncores = 6 # Change as needed

start1 = Sys.time()
# Code not running in a function #########
cl = parallel::makeCluster(ncores)
parallel::clusterExport(cl=cl, varlist=c("mat"), envir = environment())
geoms = pbapply::pblapply(seqs, function(y){sf::st_linestring(mat[c(y,y+1),c(1,2)])}, cl = cl)
parallel::stopCluster(cl)

end1 = Sys.time()

# Same code in a function ###########
func = function(mat, seqs, ncores){
  cl = parallel::makeCluster(ncores)
  parallel::clusterExport(cl=cl, varlist=c("mat"), envir = environment())
  geoms = pbapply::pblapply(seqs, function(y){sf::st_linestring(mat[c(y,y+1),c(1,2)])}, cl = cl)
  parallel::stopCluster(cl)
  return(geoms)
}

start2 = Sys.time()
res = func(mat, seqs, ncores)
end2 = Sys.time()


# Compare Results 
difftime(end1, start1, units = "s")
difftime(end2, start2, units = "s")

比较时间后，函数中代码的速度明显降低

difftime(end1, start1, units = "s")
Time difference of 41.98823 secs
difftime(end2, start2, units = "s")
Time difference of 114.983 secs

函数内部的并行R代码比外部的慢

0 个答案: