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我想在以下代码中并行化 while 循环：

work <- function(n) {
  # Do some intensive work (e.g explore a graph starting at n).
  # After this, we don't need to execute work() on nodes in excluding.
  # (e.g exclude could be the nodes explored/reached from n)
  # n is just an example. exclude can be a potentially large set.
  Sys.sleep(2)
  exclude <- c(n, sample(nodes, rbinom(1, length(nodes), 0.5)))
  return(exclude)
}

nodes <- 1:1e3

#Order of execution doesn't matter
nodes <- sample(nodes)

#parallelize this loop
while(length(nodes) > 0) {
  n <- nodes[1]
  exclude <- work(n)
  nodes <- setdiff(nodes, exclude)
}

是否在排除节点上执行并不重要work()，但我们希望尽量减少此类实例。上面 while 循环的目标是尽可能少地运行 work()

这不是一个尴尬的并行计算，所以我不知道如何parLapply直接使用。可以使用主从框架，但我不知道任何多核编程（在 Windows 上）。

作为一个具体示例，您可以将其work(n)视为graph_exploration(n)（查找所有连接到 的节点的函数n）和excluden 的连通分量中的节点。最终目标是从每个连接的组件中找到一个节点。您希望graph_exploration(n)尽可能少地运行，因为这是一项昂贵的操作。

4

1 回答 1

2

米赫尔，

这是一个建议的解决方案。

前言：

这里的核心问题（据我了解）是在进行work()数字运算时解除阻塞 while 循环。本质上，您希望循环不被阻塞，只要有资源可以启动更多的work()调用和处理。好吧怎么办？好吧，我的建议是你使用future包。

下面的示例实质上work()为每个调用创建了一个新的进程调用。但是，除非所有分配的工作进程都忙，否则调用不会阻塞 while 循环。您可以看到这一点，因为每个work()调用都有不同的进程 ID，如运行时输出中所示。

因此，每个work()独立运行，完成我们解决所有期货并返回决赛结果。

结果：

  • 连续运行时间：经过 20.61 秒
  • 并行运行时间：经过 8.22 秒

我希望这能为您指明正确的方向。

警告：您必须遍历所有节点，但它确实提高了运行时间。

机器设置：

R version 3.4.1 (2017-06-30)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows >= 8 x64 (build 9200)
[Windows 10, 8 Core Xeon, 64Gb RAM]

并行代码示例：

# Check for, and install and load required packages.
requiredPackages <-
  c("tictoc", "listenv", "future")


ipak <- function(pkg) {
  new.pkg <- pkg[!(pkg %in% installed.packages()[, "Package"])]
  if (length(new.pkg))
    install.packages(new.pkg, dependencies = TRUE)
  sapply(pkg, require, character.only = TRUE)
}

ipak(requiredPackages)

work <- function(n) {
  # Do some intensive work (e.g explore a graph starting at n).
  # After this, we don't need to execute work() on nodes in exclude.
  # (e.g exclude could be the nodes explored/reached from n)
  # n is just an example. exclude can be a potentially large set.
  Sys.sleep(2) # sample(.5:5))
  exclude <- n
  return(exclude)
}

plan(multiprocess, workers = 4L)
#plan(sequential)

nodesGraph  <- 1:10
nodesGraph  <- sample(nodesGraph)
nodesCount  <- length(nodesGraph)
resultsList <- listenv()

tic()
while ( nodesCount > 0 ) {
  n <- nodesGraph[[nodesCount]]
  ## This is evaluated in parallel and will only block
  ## if all workers are busy.
  resultsList[[nodesCount]] %<-% {
      list( exclude = work(n), 
            iteration = length(nodesGraph), 
            pid = Sys.getpid())
  }

  nodesGraph <- setdiff(nodesGraph, nodesGraph[[nodesCount]] )
  cat("nodesGraph",nodesGraph,"\n")
  cat("nodesCount",nodesCount,"\n")
  nodesCount = nodesCount - 1
}
toc()

## Resolve all futures (blocks if not already finished)
resultsList <- as.list(resultsList)
str(resultsList)

并行运行时输出：

> source('<hidden>/dev/stackoverflow/47230384/47230384v5.R')
nodesGraph 2 5 8 4 6 10 7 1 9 
nodesCount 10 
nodesGraph 2 5 8 4 6 10 7 1 
nodesCount 9 
nodesGraph 2 5 8 4 6 10 7 
nodesCount 8 
nodesGraph 2 5 8 4 6 10 
nodesCount 7 
nodesGraph 2 5 8 4 6 
nodesCount 6 
nodesGraph 2 5 8 4 
nodesCount 5 
nodesGraph 2 5 8 
nodesCount 4 
nodesGraph 2 5 
nodesCount 3 
nodesGraph 2 
nodesCount 2 
nodesGraph  
nodesCount 1 
8.22 sec elapsed
List of 10
 $ :List of 3
  ..$ exclude  : int 2
  ..$ iteration: int 1
  ..$ pid      : int 10692
 $ :List of 3
  ..$ exclude  : int 5
  ..$ iteration: int 2
  ..$ pid      : int 2032
 $ :List of 3
  ..$ exclude  : int 8
  ..$ iteration: int 3
  ..$ pid      : int 16356
 $ :List of 3
  ..$ exclude  : int 4
  ..$ iteration: int 4
  ..$ pid      : int 7756
 $ :List of 3
  ..$ exclude  : int 6
  ..$ iteration: int 5
  ..$ pid      : int 10692
 $ :List of 3
  ..$ exclude  : int 10
  ..$ iteration: int 6
  ..$ pid      : int 2032
 $ :List of 3
  ..$ exclude  : int 7
  ..$ iteration: int 7
  ..$ pid      : int 16356
 $ :List of 3
  ..$ exclude  : int 1
  ..$ iteration: int 8
  ..$ pid      : int 7756
 $ :List of 3
  ..$ exclude  : int 9
  ..$ iteration: int 9
  ..$ pid      : int 10692
 $ :List of 3
  ..$ exclude  : int 3
  ..$ iteration: int 10
  ..$ pid      : int 2032

顺序运行时输出

> source('<hidden>/dev/stackoverflow/47230384/47230384v5.R')
nodesGraph 6 2 1 9 4 8 10 7 3 
nodesCount 10 
nodesGraph 6 2 1 9 4 8 10 7 
nodesCount 9 
nodesGraph 6 2 1 9 4 8 10 
nodesCount 8 
nodesGraph 6 2 1 9 4 8 
nodesCount 7 
nodesGraph 6 2 1 9 4 
nodesCount 6 
nodesGraph 6 2 1 9 
nodesCount 5 
nodesGraph 6 2 1 
nodesCount 4 
nodesGraph 6 2 
nodesCount 3 
nodesGraph 6 
nodesCount 2 
nodesGraph  
nodesCount 1 
20.61 sec elapsed
List of 10
 $ :List of 3
  ..$ exclude  : int 6
  ..$ iteration: int 1
  ..$ pid      : int 12484
 $ :List of 3
  ..$ exclude  : int 2
  ..$ iteration: int 2
  ..$ pid      : int 12484
 $ :List of 3
  ..$ exclude  : int 1
  ..$ iteration: int 3
  ..$ pid      : int 12484
 $ :List of 3
  ..$ exclude  : int 9
  ..$ iteration: int 4
  ..$ pid      : int 12484
 $ :List of 3
  ..$ exclude  : int 4
  ..$ iteration: int 5
  ..$ pid      : int 12484
 $ :List of 3
  ..$ exclude  : int 8
  ..$ iteration: int 6
  ..$ pid      : int 12484
 $ :List of 3
  ..$ exclude  : int 10
  ..$ iteration: int 7
  ..$ pid      : int 12484
 $ :List of 3
  ..$ exclude  : int 7
  ..$ iteration: int 8
  ..$ pid      : int 12484
 $ :List of 3
  ..$ exclude  : int 3
  ..$ iteration: int 9
  ..$ pid      : int 12484
 $ :List of 3
  ..$ exclude  : int 5
  ..$ iteration: int 10
  ..$ pid      : int 12484
于 2017-11-18T05:53:22.190 回答