Question

您好我正在研究并行计算的数据。我希望看到每个k时间的总执行时间并存储到数据帧中。但是当我看到结果存储在列表时间表中时，经过的时间是负值，这对我来说没有意义。我的代码出了什么问题？

library(doParallel)
library(doRNG)
cl <- makeCluster(3); 
registerDoParallel(cl)

registerDoRNG(seed = 11)
tstart_par = proc.time()
time_chain_par = matrix(0, 10, 2)
timetable <- foreach(k=1:10, .options.RNG=11) %dorng%{
    runif(10,0,1)
    runif(10,0,1)
    runif(10,0,1)
    etime_par = proc.time() - tstart_par
}



> timetable
[[1]]
 user    system   elapsed 
  -42.559    -8.004 -1483.762 

[[2]]
 user    system   elapsed 
  -43.179    -8.038 -1484.064 

[[3]]
 user    system   elapsed 
  -42.996    -8.065 -1484.305 

[[4]]
 user    system   elapsed 
  -42.558    -8.004 -1483.761 

[[5]]
     user    system   elapsed 
  -43.178    -8.038 -1484.063 

[[6]]
 user    system   elapsed 
  -42.558    -8.004 -1483.761 

[[7]]
     user    system   elapsed 
  -42.996    -8.064 -1484.304 

[[8]]
     user    system   elapsed 
  -43.178    -8.038 -1484.062 

[[9]]
     user    system   elapsed 
  -42.558    -8.004 -1483.760 

[[10]]
     user    system   elapsed 
  -43.177    -8.038 -1484.061

Answer 1

proc.time()在群集中重置，因此您要比较主R进程运行的时长与新创建的群集运行的时间，并给出负值。在这种情况下最好使用Sys.time()函数，它比较中央时钟。因此，要计算从启动并行流程到每个群集完成处理的时间：

library(doParallel)
library(doRNG)
cl <- makeCluster(3); 
registerDoParallel(cl)

registerDoRNG(seed = 11)
tstart_par = Sys.time()

timetable <- foreach(k=1:10, .options.RNG=11) %dorng%{
  runif(10,0,1)
  runif(10,0,1)
  runif(10,0,1)
  etime_par = Sys.time() - tstart_par
}
stopCluster(cl)

> timetable
[[1]]
Time difference of 2.897783 secs

[[2]]
Time difference of 2.897755 secs

[[3]]
Time difference of 2.899669 secs

[[4]]
Time difference of 2.899365 secs

[[5]]
Time difference of 2.899258 secs

[[6]]
Time difference of 2.900039 secs

[[7]]
Time difference of 2.901185 secs

[[8]]
Time difference of 2.901083 secs

[[9]]
Time difference of 2.901798 secs

[[10]]
Time difference of 2.902622 secs

或者每个单独的集群运行多长时间你可以使用proc.time（），但是把它放在函数内部（我增加了样本数量，这样它可以运行更长时间，你可以看到差异）：

cl <- makeCluster(3); 
registerDoParallel(cl)

registerDoRNG(seed = 11)

time_chain_par = matrix(0, 10, 2)
timetable <- foreach(k=1:10, .options.RNG=11) %dorng%{
  tstart_par = proc.time()
  runif(1000000,0,1)
  etime_par = proc.time() - tstart_par
}
stopCluster(cl)

timetable
[[1]]
   user  system elapsed 
  0.148   0.008   0.307 

[[2]]
   user  system elapsed 
  0.145   0.008   0.249 

[[3]]
   user  system elapsed 
  0.145   0.008   0.325 

[[4]]
   user  system elapsed 
  0.062   0.002   0.130 

[[5]]
   user  system elapsed 
  0.062   0.002   0.118 

[[6]]
   user  system elapsed 
  0.061   0.003   0.110 

[[7]]
   user  system elapsed 
  0.062   0.002   0.110 

[[8]]
   user  system elapsed 
  0.061   0.003   0.149 

[[9]]
   user  system elapsed 
  0.062   0.002   0.117 

[[10]]
   user  system elapsed 
  0.097   0.003   0.169

并联计算并存储R中的经过时间负值

1 个答案: