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我正在尝试抓取并解析以下 RSS 提要http://www.huffingtonpost.com/rss/liveblog/liveblog-1213.xml我查看了有关 R 和 XML 的其他查询，但未能取得任何进展关于我的问题。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”（全文清理）。我曾尝试在 R 中使用 xml 包，但承认我有点新手（几乎没有使用 XML 的经验，但有一些 R 经验）。我正在处理的代码是：

 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，每当我试图操纵和试图操纵“标题”或“描述”信息时，我都会不断地出错。任何帮助解决此代码，将不胜感激。

谢谢，托马斯

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1 回答 1

11

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、发布日期和描述）

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)
于 2012-11-20T12:55:14.720 回答