Question

我有一个XML文件，其中包含父母级别，我需要将其转换为数据帧。

以下是xml文件的示例：

<Header>
    <Response>
        <Response-type>
            1
        </Response-type>
        <CODE>
            1
        </CODE>
    </Response>
    <Identification>
        <Request>
            <Request-name>
                Testing
            </Request-name>
            <Request-time>
                <Year>
                    2015
                </Year>
                <Month>
                    December
                </Month>
                <Time>
                    <Hour>
                        1
                    </Hour>
                    <Minute>
                        20
                    </Minute>
                </Time>
            </Request-time>
        </Request>
    </Identification>
</Header>

我尝试将XML文件转换为列表列表，如下所示：

library(XML)
xml <- xmlTreeParse("myfile.xml", useInternalNodes = TRUE)
xml_list <- xmlToList(xml)

当我尝试将列表转换为数据帧时，会出现问题，如下所示：

as.data.frame(t(as.data.frame(xml_list)))

我得到以下错误：

Error in (function (..., row.names = NULL, check.rows = FALSE, check.names = TRUE,  : 
  arguments imply differing number of rows: 1, 0

Answer 1

这是使用xml2库的解决方案。我假设每个标头节点只有一个响应和标识节点代码中的注释解释了一步一步的过程：

library(xml2)
page<-read_xml("<Header>
    <Response>
      <Response-type> 1  </Response-type>
         <CODE> 1 </CODE>
      </Response>
      <Identification>
         <Request>
           <Request-name> Testing </Request-name>
           <Request-time>
             <Year>  2015 </Year>
             <Month> December </Month>
             <Time>  <Hour> 1 </Hour> <Minute> 20 </Minute></Time>
           </Request-time>
         </Request>
      </Identification> </Header>")

#Find all nodes, (assuming a single node in the document)
all<-xml_find_all(page, '//*')

#Goal is to find all the nodes at the bottom of hierarchy
#Find the children for each node 
children<-lapply(all, xml_children)
#determine the number of children on each node
#and check for the number of children if ==0 then this is a leaf node
nochildren<-sapply(children, length)==0

#titles of each leaf
names<-xml_name(all[nochildren])
#values of each leaf
values<-trimws(xml_text(all[nochildren]))

#build dataframe
df<-data.frame(t(values))
names(df)<-names

使用R将不相等的XML节点转换为数据帧

1 个答案: