xml – R从data.frame获取列名

发布时间：2020-12-15 23:57:11 所属栏目：百科来源：网络整理

导读：我有以下代码： install.packages("XML")library(XML)install.packages("plyr")library(plyr)feed - "http://feeds.reuters.com/Reuters/worldNews?format=xml"reuters-xmlToList(feed)data - lapply(reuters[[1]][names(reuters[[1]])=="item"],data.frame)

我有以下代码：

install.packages("XML")
library(XML)
install.packages("plyr")
library(plyr)

feed <- "http://feeds.reuters.com/Reuters/worldNews?format=xml"
reuters<-xmlToList(feed)
data <- lapply(reuters[[1]][names(reuters[[1]])=="item"],data.frame)

data

输出所有数据.

我怎样才能获得所有数据的标题？

我试过这个名字(数据),但它只输出“item”“item”“item”.

您有一个data.frames列表.您可以将它们绑定在一起：

> names(do.call(rbind.data.frame,data))
[1] "title"           "link"            "description"     "category.text"  
[5] "category..attrs" "pubDate"         "guid.text"       "guid..attrs"    
[9] "origLink"

data1 <- do.call(rbind.data.frame,data)
> head(data1$title)
[1] Niger says will repatriate its illegal migrants from Algeria     
[2] Twin bombing near Kurdish party office in north Iraq kills 30    
[3] Suicide bomber kills four soldiers in Pakistan's tribal northwest
[4] Sisi keeps Egyptian premier to fix economy after turmoil         
[5] Kosovo's Thaci has tough job to form new cabinet,keep promises  
[6] Libyan Supreme Court rules PM's election unconstitutional        
25 Levels: Niger says will repatriate its illegal migrants from Algeria ...

如果你只想要标题

xData <- xmlParse(feed)
> head(xpathSApply(xData,"//title",xmlValue))
[1] "Reuters: World News"                                                 
[2] "Reuters: World News"                                                 
[3] "South Africa platinum strike talks in crucial final day of mediation"
[4] "Africa's sports bars,TV shacks step up security for World Cup"      
[5] "Niger says will repatriate its illegal migrants from Algeria"        
[6] "Twin bombing near Kurdish party office in north Iraq kills 30"

（编辑：李大同）

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