使用并行化来用R刮取网页

Question

我试图刮掉大量的网页,以便以后分析它们.由于URL的数量巨大,我决定使用该parallel包XML.

具体来说,我正在使用htmlParse()函数from XML,它在使用时工作正常sapply,但在使用时会生成类HTMLInternalDocument的空对象parSapply.

url1<- "http://forums.philosophyforums.com/threads/senses-of-truth-63636.html"
url2<- "http://forums.philosophyforums.com/threads/the-limits-of-my-language-impossibly-mean-the-limits-of-my-world-62183.html"
url3<- "http://forums.philosophyforums.com/threads/how-language-models-reality-63487.html"

myFunction<- function(x){
cl<- makeCluster(getOption("cl.cores",detectCores()))
ok<- parSapply(cl=cl,X=x,FUN=htmlParse)
return(ok)
}

urls<- c(url1,url2,url3)

#Works
output1<- sapply(urls,function(x)htmlParse(x))
str(output1[[1]])
> Classes 'HTMLInternalDocument', 'HTMLInternalDocument', 'XMLInternalDocument', 'XMLAbstractDocument', 'oldClass' <externalptr>
output1[[1]]


#Doesn't work
myFunction<- function(x){
cl<- makeCluster(getOption("cl.cores",detectCores()))
ok<- parSapply(cl=cl,X=x,FUN=htmlParse)
stopCluster(cl)
return(ok)
}

output2<- myFunction(urls)
str(output2[[1]])
> Classes 'HTMLInternalDocument', 'HTMLInternalDocument', 'XMLInternalDocument', 'XMLAbstractDocument', 'oldClass' <externalptr>
output2[[1]]
#empty

谢谢.

Answer 1

您可以使用getURIAsynchronousRcurl包,允许调用者指定多个URI同时下载.

library(RCurl)
library(XML)
get.asynch <- function(urls){
  txt <- getURIAsynchronous(urls)
  ## this part can be easily parallelized 
  ## I am juste using lapply here as first attempt
  res <- lapply(txt,function(x){
    doc <- htmlParse(x,asText=TRUE)
    xpathSApply(doc,"/html/body/h2[2]",xmlValue)
  })
}

get.synch <- function(urls){
  lapply(urls,function(x){
    doc <- htmlParse(x)
    res2 <- xpathSApply(doc,"/html/body/h2[2]",xmlValue)
    res2
  })}

这里有一些100 urls的基准测试,你将解析时间除以2倍.

library(microbenchmark)
uris = c("http://www.omegahat.org/RCurl/index.html")
urls <- replicate(100,uris)
microbenchmark(get.asynch(urls),get.synch(urls),times=1)

Unit: seconds
             expr      min       lq   median       uq      max neval
 get.asynch(urls) 22.53783 22.53783 22.53783 22.53783 22.53783     1
  get.synch(urls) 39.50615 39.50615 39.50615 39.50615 39.50615     1