Question

我的代码需要花费大量时间执行：

dataRaw<-pblapply(femme,function (x) {

  article<-user(x,date=FALSE,weight=FALSE)
  names<-rep(x,length(article))  
  result<-matrix(c(names,article),ncol=2)

})
dataRaw<-do.call(rbind,dataRaw)
dataRaw[,3]<-vector(length=length(dataRaw[,2]))
dataRaw[,3]<-pbapply(dataRaw,1,function(x){
  Rprof(filename = "profile.out")
  revisions<-revisionsPage(x[2])
  rank<-rankingContrib(revisions,50)
  rank<-rank$contrib
  x[1] %in% rank
  Rprof(NULL)
})
result<-as.vector(dataRaw[dataRaw$ranking==TRUE,2])

laching summaryRprof函数，它给我这个

$by.self
                        self.time self.pct total.time total.pct
".Call"                      0.46    95.83       0.46     95.83
"as.data.frame.numeric"      0.02     4.17       0.02      4.17

$by.total
                             total.time total.pct self.time self.pct
"FUN"                              0.48    100.00      0.00     0.00
"pbapply"                          0.48    100.00      0.00     0.00
".Call"                            0.46     95.83      0.46    95.83
"<Anonymous>"                      0.46     95.83      0.00     0.00
"GET"                              0.46     95.83      0.00     0.00
"request_fetch"                    0.46     95.83      0.00     0.00
"request_fetch.write_memory"       0.46     95.83      0.00     0.00
"request_perform"                  0.46     95.83      0.00     0.00
"revisionsPage"                    0.46     95.83      0.00     0.00
"as.data.frame.numeric"            0.02      4.17      0.02     4.17
"as.data.frame"                    0.02      4.17      0.00     0.00
"data.frame"                       0.02      4.17      0.00     0.00
"rankingContrib"                   0.02      4.17      0.00     0.00

$sample.interval
[1] 0.02

$sampling.time
[1] 0.48

看来它是＆＃34; .Call＆＃34;占用所有机器时间的功能。这是什么。电话入口？

解释分析文件（Rprof）

0 个答案: