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"
data <- ldply(xmlToList(feed), data.frame)但是,它给出了以下错误:
Error in data.frame(title = "Reuters: World News", link =
"http://www.reuters.com", : arguments imply differing number of
rows: 1, 3, 2为什么我不能加载这个XML (但我可以加载其他XML,比如www.w3schools.com/XQuery/books.xml)?
发布于 2014-06-06 20:09:22
还有一个函数xmlToDataFrame
library(XML)
feed <- "http://feeds.reuters.com/Reuters/worldNews?format=xml"
(data <- xmlToDataFrame(xmlParse(feed)["/rss/channel/item"]))
# dplyr::glimpse(data)
# Variables:
# $ title (fctr) More than 60 migrants drown in boat sinking off Yemen:...
# $ link (fctr) http://feeds.reuters.com/~r/Reuters/worldNews/~3/p08tv...
# $ description (fctr) GENEVA (Reuters) - At least 60 African migrants and tw...
# $ category (fctr) worldNews, worldNews, worldNews, worldNews, worldNews,...
# $ pubDate (fctr) Fri, 06 Jun 2014 19:18:12 GMT, Fri, 06 Jun 2014 19:01:...
# $ guid (fctr) http://www.reuters.com/article/2014/06/06/us-yemen-mig...
# $ origLink (fctr) http://reuters.us.feedsportal.com/c/35217/f/654198/s/3...发布于 2014-06-06 20:03:02
我将猜测,您只需要对结果中的所有"item“节点进行data.frames。如果是这样的话
feed <- "http://feeds.reuters.com/Reuters/worldNews?format=xml"
reuters<-xmlToList(feed)
lapply(reuters[[1]][names(reuters[[1]])=="item"], data.frame)应该能起作用。
https://stackoverflow.com/questions/24089644
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