mclapply用户时间大于经过时间 [英] mclapply user time larger than elapsed time

问题描述

我正在尝试使用R中parallel软件包的mclapply功能.该函数通过计算对数似然距离将值分配给序列矩阵，这是CPU密集型操作.

I am trying to use mclapply function of the parallel package in R. The function is assigning the values to sequence matrix by calculating log likelihood distance - a CPU-intensive operation.

产生的system.time值令人困惑:

> system.time(mclapply(worksample,function(x){p_seqi_modj(x,worksample[[1]],c(1:17))}))
   user  system elapsed 
 29.339   1.242  18.581 

我认为elapsed表示合计时间(user + system).在这种情况下，以上结果意味着什么?我应该在什么时候定向自己?我无与伦比的版本在user中花费的时间更少，而在elapsed中花费的时间更多.

I thought that elapsed means aggregated time (user+system). What does the above result mean in this case and to what time should I orient myself? My unparallelized version takes less in user time and much more in elapsed.

推荐答案

帮助页面?system.time表示该函数返回的值是proc_time类的对象，我们应该参考?proc.time.在那里，我们了解到用户时间是

The help page ?system.time says the value returned by the function is an object of class proc_time, and that we should consult ?proc.time. There we learn that user time is

cumulative sum of user and system times of any child processes

因此您的任务在每个内核上花费了大约15秒(mclapply默认使用2个内核，请参见mc.cores参数).

so your task has spent about 15s on each core (mclapply defaults to using 2 cores, see the mc.cores argument).

实际上，我们早先在帮助页面中看到proc.time()返回五个将流程时间和子时间分开的元素，并且打印中使用的summary方法将用户时间和系统时间折叠为流程时间+子时间.还有更多可用信息.

Actually, we see earlier in the help page that proc.time() returns five elements that separate the process and child times, and that the summary method used in printing collapses the user and system time into process + child times, so there is a bit more information available.

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