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)
}
work()是否在排除的节点上执行并不重要，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()
这不是一个令人尴尬的并行计算，所以我不知道如何直接使用parLapply。可以使用主从框架，但我不知道(在Windows上)有任何用于多核编程的框架。
作为一个具体的例子，你可以把work(n)想象成graph_exploration(n)(查找连接到n的所有节点的函数)，把exclude想象成n的连接组件中的节点。最终目标是从每个连接组件中找到一个节点。您希望尽可能少地运行graph_exploration(n)，因为这是一个昂贵的操作。
Miheer，
这是一个建议的解决方案。
序言：
这里的核心问题(据我所知)是在work()进行数字运算的同时，解除while循环的阻塞。从本质上讲，只要有资源可以启动更多的work()调用和处理，就希望循环不被阻塞。好的，怎么做？好吧，我的建议是你使用未来的套餐。
下面的示例实质上为每个调用创建一个新的进程调用work()。但是，除非所有分配的工作进程都很忙，否则调用不会阻塞while循环。您可以看到这一点，因为每个work()调用都有一个不同的进程id，如运行时输出中所示。
因此，每个work()都是独立运行的，为了完成，我们解析了所有的未来并返回最终结果。
结果：
- 顺序运行时间：运行了20.61秒
- 并行运行时：运行了8.22秒
我希望这能给你指明正确的方向。
注意事项： 您必须运行所有节点，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