解析大型非结构化xml文件

I have a very complicated xml file need to parse and present in dataframe format in R. The structure may similar to the following example. The nodes are not paralleled.

我有一个非常复杂的xml文件需要解析并以r的dataframe格式呈现。节点不是并行的。

<Root>
  <A>
   <info1>a</info1>
     <child>
       <info2>b</info2>
       <info3>c</info3>
       <info4>d</info4>
     </child>
   <info5>e</info5>
  </A>
  <B>
   <info6>f</info6>
   <info7>g</info7>
  </B>
</Root>

I come up some code to parse the file:

我找到一些代码来解析这个文件:

doc <- xmlParse(file="sample.xml", useInternal = TRUE)
rootnode <- xmlRoot(doc)
df1<-xmlToDataFrame(nodes=getNodeSet(rootnode, "//Root/A"))
df2<-xmlToDataFrame(nodes=getNodeSet(rootnode, "//Root/B"))
Final<-cbind.data.frame(df1,df2, all=TRUE)

The result returned as: (all the value form node were shrink together)

返回的结果为:(所有值表单节点一起收缩)

info1 child info5 info6 info7
  a    bcd    e     f     g

However, the ideal result I want is:

但是，我想要的理想结果是:

info1 info2 info3 info4 info5 info6 info7
  a     b     c     d     e     f     g

Because there are large number of nodes in the xml file similar to the situation above, it is not wise to manually manipulate the dataframe.
I also try to change the path statement to "//Root/A/child", then all the value under node A and node B will be missed. Does anyone could offer the solution to this problem. Thanks in advance.

因为xml文件中有大量的节点与上面的情况类似，所以手工操作dataframe是不明智的。我还尝试将path语句更改为“//Root/A/child”，然后会遗漏节点A和节点B下的所有值。有没有人能提出这个问题的解决方案?提前谢谢。

3 个解决方案

#1

One can try xmlToList and unlist to reduce xml data in named vector format. The names can be changed using gsub to match OP's expectations as:

可以尝试xmlToList和unlist来减少命名向量格式的xml数据。使用gsub可以更改名称，以符合OP的期望:

library(XML)
result <- unlist(xmlToList(xmlParse(xml)))
#Change the name to refer only child 
names(result) <- gsub(".*\\.(\\w+)$","\\1", names(result))
result 
# info1 info2 info3 info4 info5 info6 info7 
# "a"   "b"   "c"   "d"   "e"   "f"   "g"

Data:

数据:

xml <- "<Root>
  <A>
  <info1>a</info1>
  <child>
  <info2>b</info2>
  <info3>c</info3>
  <info4>d</info4>
  </child>
  <info5>e</info5>
  </A>
  <B>
  <info6>f</info6>
  <info7>g</info7>
  </B>
  </Root>"

#2

In a less structured XML, it is better to do the following:

在非结构化的XML中，最好做到以下几点:

library(XML)
Final <- data.frame(xmlToList(rootnode), recursive = T, use.names = T)

If you don't like the automatically set column names, you can simply do use.names = F and set your own names.

如果您不喜欢自动设置列名，您可以使用。names = F并设置您自己的列名。

#3

Match the nodes using starts-with()

使用starts-with()匹配节点

> doc = xmlParse(xml)
> xpathSApply(doc, "//*[starts-with(name(), 'info')]", xmlValue)
[1] "a" "b" "c" "d" "e" "f" "g"
> xpathSApply(doc, "//*[starts-with(name(), 'info')]", xmlName)
[1] "info1" "info2" "info3" "info4" "info5" "info6" "info7"

所以

query <- "//*[starts-with(name(), 'info')]"
setNames(
    xpathSApply(doc, query, xmlValue),
    xpathSApply(doc, query, xmlName)
)

#1