Question

我正在尝试抓取并解析以下RSS提要http://www.huffingtonpost.com/rss/liveblog/liveblog-1213.xml我已经查看了有关R和XML的其他查询，并且无法在我的问题上取得任何进展。每个条目的xml代码

        <item>
     <title><![CDATA[Five Rockets Intercepted By Iron Drone Systems Over Be'er Sheva]]></title>
     <link>http://www.huffingtonpost.co.uk/2012/11/15/tel-aviv-gaza-rocket_n_2138159.html#2_five-rockets-intercepted-by-iron-drone-systems-over-beer-sheva</link>
     <description><![CDATA[<a href="http://www.haaretz.com/news/diplomacy-defense/live-blog-rockets-strike-tel-aviv-area-three-israelis-killed-in-attack-on-south-1.477960" target="_hplink">Haaretz reports</a> that five more rockets intercepted by Iron Dome systems over Be'er Sheva. In total, there have been 274 rockets fired and 105 intercepted. The IDF has attacked 250 targets in Gaza.]]></description>
     <guid>http://www.huffingtonpost.co.uk/2012/11/15/tel-aviv-gaza-rocket_n_2138159.html#2_five-rockets-intercepted-by-iron-drone-systems-over-beer-sheva</guid>
     <pubDate>2012-11-15T12:56:09-05:00</pubDate>
     <source url="http://huffingtonpost.com/rss/liveblog/liveblog-1213.xml">Huffingtonpost.com</source>
  </item>

对于每个条目/帖子我想记录“日期”（pubDate），“标题”（标题），“描述”（全文清除）。我曾尝试在R中使用xml包，但承认我是一个新手（很少有没有使用XML的经验，但有些R经验）。我正在处理的代码是无处可去的：

 library(XML)

 xml.url <- "http://www.huffingtonpost.com/rss/liveblog/liveblog-1213.xml"

 # Use the xmlTreePares-function to parse xml file directly from the web

 xmlfile <- xmlTreeParse(xml.url)

# Use the xmlRoot-function to access the top node

xmltop = xmlRoot(xmlfile)

xmlName(xmltop)

names( xmltop[[ 1 ]] )

  title          link   description      language     copyright 
  "title"        "link" "description"    "language"   "copyright" 
 category     generator          docs          item          item 
  "category"   "generator"        "docs"        "item"        "item"

然而，每当我试图操纵并试图操纵“标题”或“描述”信息时，我都会不断地收到错误。任何帮助解决此代码的帮助都将非常受欢迎。

谢谢，托马斯

Answer 1

我正在使用优秀的Rcurl库和xpathSApply

这是脚本为您提供3个列表（标题，pubdates和描述）

library(RCurl)
library(XML)
xml.url <- "http://www.huffingtonpost.com/rss/liveblog/liveblog-1213.xml"
script  <- getURL(xml.url)
doc     <- xmlParse(script)
titles    <- xpathSApply(doc,'//item/title',xmlValue)
descriptions    <- xpathSApply(doc,'//item/description',xmlValue)
pubdates <- xpathSApply(doc,'//item/pubDate',xmlValue)

使用XML packagin R解析RSS提要

1 个答案: