如何阻止R离开僵尸进程 [英] How to stop R from leaving zombie processes behind
问题描述
library（doMC）
library（doParallel）
registerDoMC（4）
timing< - system.time（fitall < - foreach（i = 1：1000，.combine =c）％dopar％{
print（i）
}）
我启动 R 和看看进程表：
>系统（ps -efl）
F S UID PID PPID C PRI NI ADDR SZ WCHAN STIME TTY TIME CMD
4 S chbr 1 0 5 80 0 - 21399 wait 10:58？ 00:00:00 / usr / local / lib / R / bin / exec / R --no-save --no-restore
0 S chbr 9 1 0 80 0 - 1113 wait 10:58？ 00:00:00 sh -c ps -efl
0 R chbr 10 9 0 80 0 - 4294 - 10:58？ 00:00:00 ps -efl
如果我使用上面提到的简单循环 doMC 或 doParallel 在后面留下一个僵尸进程。输出 ps -efl 运行循环后：
>系统（ps -efl）
F S UID PID PPID C PRI NI ADDR SZ WCHAN STIME TTY TIME CMD
4 S chbr 1 0 4 80 0 - 25256 wait 11:00？ 00:00:00 / usr / local / lib / R / b
1 Z chbr 10 1 0 80 0 - 0 exit 11:00？ 00:00:00 [R]< defunct>
0 S chbr 12 1 0 80 0 - 1113等待11:00？ 00:00:00 sh -c ps -efl
0 R chbr 13 12 0 80 0 - 4294 - 11:00？ 00:00:00 ps -efl
如果我不重复 registerDoMC（4）再次没有额外的僵尸进程被创建。但是，如果我发行 registerDoMC（4），则会创建一个额外的僵尸进程：
>系统（ps -efl）
F S UID PID PPID C PRI NI ADDR SZ WCHAN STIME TTY TIME CMD
4 S chbr 1 0 0 80 0 - 25554等待11:00？ 00:00:01 / usr / local / lib / R / b
1 Z chbr 21 1 0 80 0 - 0 exit 11:02？ 00:00:00 [R]< defunct>
1 z chbr 22 1 0 80 0 - 0 exit 11:02？ 00:00:00 [R]< defunct>
0 S chbr 26 1 0 80 0 - 1113等待11:03？ 00:00:00 sh -c ps -efl
0 R chbr 27 26 0 80 0 - 4294 - 11:03？ 00:00:00 ps -efl
这就是我认为它可以是 doMC 这是做不应该做的事情。如果doMC造成这种情况，有没有办法阻止 doMC 离开僵尸进程？ （ stopCluster（）不起作用，因为没有创建群集。）
> sessionInfo（）
R开发中（unstable）（2014-08-16 r66404）
平台：x86_64-unknown-linux-gnu（64位）
语言环境：$ LC_CTYPE = en_IE.UTF-8 LC_NUMERIC = C
[3] LC_TIME = en_IE.UTF-8 LC_COLLATE = en_IE.UTF-8
[5] LC_MONETARY = en_IE.UTF- LC_MESSAGES = en_IE.UTF-8
[7] LC_PAPER = en_IE.UTF-8 LC_NAME = C
[9] LC_ADDRESS = C LC_TELEPHONE = C
[11] LC_MEASUREMENT = en_IE.UTF -8 LC_IDENTIFICATION = C
附加的基本软件包：
[1] parallel stats graphics grDevices utils数据集方法
[8] base
其他附加软件包：
[1] doParallel_1.0.8 doMC_1.3.3 iterators_1.0.7 foreach_1.4.2
通过命名空间加载（而不是附加）：
[1] codetools_0.2-8 compiler_3.2.0
这与foreach无关或者doMC;正如Steve Weston在回答其他StackOverflow查询时指出的那样，doMC基本上只是mclapply的一个包装，你可以看到通过对mclapply的简单调用创建的僵尸进程：
<$ （并行）
mclapply（rep（5,4），rnorm）
在我的系统中，这会留下两个僵尸进程：
[richcalaway @ richcalaway-pc〜] $ ps -efl | grep defunct
1 Z 1660945517 28701 28624 0 77 0 - 0出口12:00 pts / 1 00:00:00 [R]< defunct>
1 Z 1660945517 28702 28624 0 78 0 - 0出口12:00 pts / 1 00:00:00 [r] <不存在>
0 S 1660945517 28704 28308 0 78 0 - 15306 pipe_w 12:00 pts / 2 00:00:00 grep defunct
在正常情况下，这些僵尸进程不会造成任何麻烦，并且在R会话结束时会消失。您可以通过使用doParallel和一个fork群集来避免它们，而不是使用doMC。
干杯，
首席项目经理
革命分析
Here is a little reproducible example:
library(doMC)
library(doParallel)
registerDoMC(4)
timing <- system.time( fitall <- foreach(i=1:1000, .combine = "c") %dopar% {
print(i)
})
I start up R and look at the process table:
> system("ps -efl")
F S UID PID PPID C PRI NI ADDR SZ WCHAN STIME TTY TIME CMD
4 S chbr 1 0 5 80 0 - 21399 wait 10:58 ? 00:00:00 /usr/local/lib/R/bin/exec/R --no-save --no-restore
0 S chbr 9 1 0 80 0 - 1113 wait 10:58 ? 00:00:00 sh -c ps -efl
0 R chbr 10 9 0 80 0 - 4294 - 10:58 ? 00:00:00 ps -efl
If I use the aformentioned simple for loop doMC or doParallel leave a zombie process behind. Output of ps -efl after running the loop:
> system("ps -efl")
F S UID PID PPID C PRI NI ADDR SZ WCHAN STIME TTY TIME CMD
4 S chbr 1 0 4 80 0 - 25256 wait 11:00 ? 00:00:00 /usr/local/lib/R/b
1 Z chbr 10 1 0 80 0 - 0 exit 11:00 ? 00:00:00 [R] <defunct>
0 S chbr 12 1 0 80 0 - 1113 wait 11:00 ? 00:00:00 sh -c ps -efl
0 R chbr 13 12 0 80 0 - 4294 - 11:00 ? 00:00:00 ps -efl
If I repeat the loop without issuing registerDoMC(4) again no additional zombie process gets created. However, if I issue registerDoMC(4) an additional zombie process gets created:
> system("ps -efl")
F S UID PID PPID C PRI NI ADDR SZ WCHAN STIME TTY TIME CMD
4 S chbr 1 0 0 80 0 - 25554 wait 11:00 ? 00:00:01 /usr/local/lib/R/b
1 Z chbr 21 1 0 80 0 - 0 exit 11:02 ? 00:00:00 [R] <defunct>
1 Z chbr 22 1 0 80 0 - 0 exit 11:02 ? 00:00:00 [R] <defunct>
0 S chbr 26 1 0 80 0 - 1113 wait 11:03 ? 00:00:00 sh -c ps -efl
0 R chbr 27 26 0 80 0 - 4294 - 11:03 ? 00:00:00 ps -efl
That's how I figured it could be doMC which is doing something that should not be done. If doMC is causing this is there a way to stop doMC from leaving zombie processes behind? (stopCluster() does not work as no cluster gets created in the first place.)
> sessionInfo()
R Under development (unstable) (2014-08-16 r66404)
Platform: x86_64-unknown-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_IE.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_IE.UTF-8 LC_COLLATE=en_IE.UTF-8
[5] LC_MONETARY=en_IE.UTF-8 LC_MESSAGES=en_IE.UTF-8
[7] LC_PAPER=en_IE.UTF-8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_IE.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] parallel stats graphics grDevices utils datasets methods
[8] base
other attached packages:
[1] doParallel_1.0.8 doMC_1.3.3 iterators_1.0.7 foreach_1.4.2
loaded via a namespace (and not attached):
[1] codetools_0.2-8 compiler_3.2.0
This really has nothing to do with foreach or doMC; as Steve Weston has pointed out in answer to other StackOverflow queries, doMC is essentially just a wrapper for mclapply, and you can see zombie processes created with a simple call to mclapply:
library(parallel)
mclapply(rep(5,4), rnorm)
On my system, this leaves two zombie processes:
[richcalaway@richcalaway-pc ~]$ ps -efl | grep defunct
1 Z 1660945517 28701 28624 0 77 0 - 0 exit 12:00 pts/1 00:00:00 [R] <defunct>
1 Z 1660945517 28702 28624 0 78 0 - 0 exit 12:00 pts/1 00:00:00 [R] <defunct>
0 S 1660945517 28704 28308 0 78 0 - 15306 pipe_w 12:00 pts/2 00:00:00 grep defunct
Under normal circumstances, these zombie processes won't cause any trouble, and they do disappear when the R session ends. You can avoid them by using doParallel and a fork cluster instead of using doMC.
Cheers,
Rich Calaway
Principal Program Manager
Revolution Analytics
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