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
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输出所有数据.
我怎样才能得到所有title的的data?
我试过这个,names(data)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 ...
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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"
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