R中的XML解析

Question

如何从如下XML中提取地址：

<address class=\"addressReset\"> 
    <span rel=\"v:address\"> 
        <span dir=\"ltr\">
            <span class=\"street-address\" property=\"v:street-address\">77 Yesler Way</span>, 
            <span class=\"locality\">
                <span property=\"v:locality\">Seattle</span>, 
                <span property=\"v:region\">WA</span> 
                <span property=\"v:postal-code\">98104-2530</span>
             </span> 
        </span> 
     </span> 
</address>

我想将这些值提取到变量streetAddr、City、State、Zipcode中。

我试着做：

require(XML)
data <- xmlParse(Address)
xml_data <- xmlToList(data)

但是我的xml_data非常不格式化，我不知道如何使用它。例如,

xml_data$body$address$span$span$span$text 

给出街道地址。有更好的方法来解析这个吗？我可以使用类和属性来获取我想要的值吗？

Answer 1

另一个library(XML)解决方案使用XPath。解析文本

library(XML)
xml = xmlParse(txt)

创建XPath查询

properties = c("street-address", "locality", "region", "postal-code")
queries = sprintf("//span[@property='v:%s']/text()", properties)
names(queries) = properties

在每个查询上使用xpathSApply()和xmlValue()检索结果

sapply(queries, xpathSApply, doc=xml, fun=xmlValue)

结果是(字符)矩阵。

> sapply(queries, xpathSApply, doc=xml, fun=xmlValue)
      street-address  locality      region  postal-code 
[1,] "77 Yesler Way" "Seattle"     "WA"    "98104-2530"
[2,] "88 Yesler Way" "Seattttttle" "WAAAA" "99999-2530"

Answer 2

为什么不创建一个XML树并导航呢？

library(XML)
datatree <- xmlTreeParse(Address)
topNode <- xmlRoot(datatree)
address <- xmlSApply(topNode, function(x) xmlSApply(x[[1]][[1]][[1]], xmlValue))
print(address)

Answer 3

以下是xml2的一种方法

library(xml2)
library(purrr)

txt <- '<?xml version=\"1.0\" encoding=\"UTF-8\"?>
<addresses>
<address class=\"addressReset\"> 
<span rel=\"v:address\"> 
<span dir=\"ltr\">
<span class=\"street-address\" property=\"v:street-address\">77 Yesler Way</span>, 
<span class=\"locality\">
<span property=\"v:locality\">Seattle</span>, 
<span property=\"v:region\">WA</span> 
<span property=\"v:postal-code\">98104-2530</span>
</span> 
</span> 
</span> 
</address>
<address class=\"addressReset\"> 
<span rel=\"v:address\"> 
<span dir=\"ltr\">
<span class=\"street-address\" property=\"v:street-address\">88 Yesler Way</span>, 
<span class=\"locality\">
<span property=\"v:locality\">Seattttttle</span>, 
<span property=\"v:region\">WAAAA</span> 
<span property=\"v:postal-code\">99999-2530</span>
</span> 
</span> 
</span> 
</address>
</addresses>
'

doc <- read_xml(txt)

properties <- c("v:street-address", "v:locality", "v:region", "v:postal-code")

map_df(xml_find_all(doc, "//address"), function(x) {

  data.frame(as.list(set_names(map_chr(properties, function(y) {
    xml_text(xml_find_all(x, sprintf(".//span[@property='%s']", y)))
  }), properties)), stringsAsFactors=FALSE)

})

## Source: local data frame [2 x 4]
## 
##   v.street.address  v.locality v.region v.postal.code
##              (chr)       (chr)    (chr)         (chr)
## 1    77 Yesler Way     Seattle       WA    98104-2530
## 2    88 Yesler Way Seattttttle    WAAAA    99999-2530

我想应该有父母的标签。这可能是一个错误的假设，但推断应该是简单明了的。

对于每个地址，使用命名属性提取每个跨度，将它们转换为数据帧并将它们全部绑定在一起。