Question

我正试图从global Dept stores中删除大陆/国家的Dept商店列表。我正在运行以下代码以首先获取各大洲，因为我们可以看到XML层次结构的方式使得每个大陆的国家不是该大陆的子节点。

> url<-"http://en.wikipedia.org/wiki/List_of_department_stores_by_country"
> doc = htmlTreeParse(url, useInternalNodes = T)
> nodeNames = getNodeSet(doc, "//h2/span[@class='mw-headline']")
> # For Africa
> xmlChildren(nodeNames[[1]])
$a
<a href="/wiki/Africa" title="Africa">Africa</a> 

attr(,"class")
[1] "XMLInternalNodeList" "XMLNodeList"        
> xmlSize(nodeNames[[1]])
[1] 1

我知道我可以在单独的getNodeSet命令中执行这些国家/地区，但我只是想确保我没有遗漏某些内容。是否有更智能的方法来获取每个大陆内的所有数据，然后在每个国家/地区内同时获取所有数据？

Answer 1

uisng xpath，几个路径可以与|组合分隔器。所以我用它来获得同一个列表中的contries和商店。然后我得到第二个contries列表。我使用后一个列表来分割第一个

url<-"http://en.wikipedia.org/wiki/List_of_department_stores_by_country"
library(XML)
xmltext <- htmlTreeParse(url, useInternalNodes = T)

## Here I use the combined xpath 
cont.shops <- xpathApply(xmltext, '//*[@id="mw-content-text"]/ul/li|
                                   //*[@id="mw-content-text"]/h3',xmlValue)
cont.shops<- do.call(rbind,cont.shops)                  ## from list to  vector


head(cont.shops)                  ## first element is country followed by shops
     [,1]                   
[1,] "[edit] Â Tunisia"     
[2,] "Magasin GÃƒÂ©nÃƒÂ©ral"
[3,] "Mercure Market"       
[4,] "Promogro"             
[5,] "Geant"                
[6,] "Carrefour"            
## I get all the contries in one list 
contries <- xpathApply(xmltext, '//*[@id="mw-content-text"]/h3',xmlValue)
contries <- do.call(rbind,contries)                     ## from list to  vector

    head(contries)
     [,1]                   
[1,] "[edit] Â Tunisia"     
[2,] "[edit] Â Morocco"     
[3,] "[edit] Â Ghana"       
[4,] "[edit] Â Kenya"       
[5,] "[edit] Â Nigeria"     
[6,] "[edit] Â South Africa"

现在我进行一些处理，以便使用国家/地区拆分对象。

dd <- which(cont.shops %in% contries)                   ## get the index of contries
freq <- c(diff(dd),length(cont.shops)-tail(dd,1)+1)     ## use diff to get Frequencies
contries.f <- rep(contries,freq)                        ## create the factor splitter


ll <- split(cont.shops,contries.f)

我可以查看结果：

> ll[[contries[1]]]
[1] "[edit] Â Tunisia"      "Magasin GÃƒÂ©nÃƒÂ©ral" "Mercure Market"        "Promogro"              "Geant"                
[6] "Carrefour"             "Monoprix"             
> ll[[contries[2]]]
[1] "[edit] Â Morocco"                                                         
[2] "Alpha 55, one 6-story store in Casablanca"                                
[3] "Galeries Lafayette, to open in 2011[1] within Morocco Mall, in Casablanca"

刮取分层数据

1 个答案: