Question

我从rNomads包中获取了以下代码并对其进行了一些修改。

最初运行时，我得到：

> WebCrawler(url = "www.bikeforums.net")
[1] "www.bikeforums.net"
[1] "www.bikeforums.net"

Warning message:
XML content does not seem to be XML: 'www.bikeforums.net'

以下是代码：

require("XML")

# cleaning workspace
rm(list = ls())

# This function recursively searches for links in the given url and follows every single link.
# It returns a list of the final (dead end) URLs.
# depth - How many links to return. This avoids having to recursively scan hundreds of links. Defaults to NULL, which returns everything.
WebCrawler <- function(url, depth = NULL, verbose = TRUE) {

  doc <- XML::htmlParse(url)
  links <- XML::xpathSApply(doc, "//a/@href")
  XML::free(doc)
  if(is.null(links)) {
    if(verbose) {
      print(url)
    }
    return(url)
  } else {
    urls.out <- vector("list", length = length(links))
    for(link in links) {
      if(!is.null(depth)) {
        if(length(unlist(urls.out)) >= depth) {
          break
        }
      }
      urls.out[[link]] <- WebCrawler(link, depth = depth, verbose = verbose)
    }
    return(urls.out)
  }
}


# Execution
WebCrawler(url = "www.bikeforums.net")

任何建议我做错了什么？

更新

大家好，

我开始这个赏金，因为我认为在R社区中需要这样一个可以抓取网页的功能。赢得奖金的解决方案应该显示一个带有两个参数的函数：

WebCrawler(url = "www.bikeforums.net", xpath = "\\title" )

作为输出，我希望有一个包含两列的数据框：网站链接，如果示例xpath表达式匹配具有匹配表达式的列。

我非常感谢您的回复

Answer 1

在您的函数中的links <- XML::xpathSApply(doc, "//a/@href")下插入以下代码。

links <- XML::xpathSApply(doc, "//a/@href")
links1 <- links[grepl("http", links)] # As @Floo0 pointed out this is to capture non relative links
links2 <- paste0(url, links[!grepl("http", links)]) # and to capture relative links
links <- c(links1, links2)

还要记得url为http://www......

此外，您还没有更新urls.out列表。正如你所拥有的那样，它总是一个长度与links的长度相同的空列表

R WebCrawler - XML内容似乎不是XML：

1 个答案: