R 将XML文件转换为数据帧

我想读XML 860文件

所有xml文件都具有该结构

<ip id_pac="48">
      <rodcis>48</rodcis>
      <jmeno>Andrew</jmeno>
      <prijmeni>Mazal</prijmeni>
      <titul_pred></titul_pred>
      <titul_za></titul_za>
      <dat_dn format="D">1999-06-21</dat_dn>
      <dat_de format="D"></dat_de>
      <sex>M</sex>
      <rod_prijm></rod_prijm>
      <a typ="1">
        <dat_od format="D">2020-09-17</dat_od>
      </a>
    </ip>

48
安得烈
马扎尔
1999-06-21
M
2020-09-17

我想得到一个数据框，其中一列是“rodcis”（在本例中为48），第二列是“dat_od”（在本例中为2020-09-17）

我正在尝试这个

获取所有XML文件

---

files下面是一个使用xml2包的可能解决方案。请参阅代码中的注释，以了解分步指南

library(xml2)
library(dplyr)

#loop through the file list with lapply
dfs <-lapply(files, function(file) {
   #read file
   page <- read_xml(file)
   
   #get id for each file
   id <- xml_find_first(out, "//ip") %>% xml_attr("id_pac") 
   
   #get information from each the requested nodes
   rodcis <- xml_find_first(out, ".//rodcis") %>% xml_text()
   datod <- xml_find_first(out, ".//dat_od") %>% xml_text()
   
   #make data frame of results for each file
   data.frame(id, rodcis, datod)
})

#combine all results into 1 data frame
answer <- bind_rows(dfs)

库（xml2）
图书馆（dplyr）
#使用lappy循环浏览文件列表
dfs
    out <- lapply(files, xmlParse)

    dataframe <- do.call(rbind, lapply(out, function(x) rootnode[[1]][[2]], rootnode[[2]][[1]]))

library(xml2)
library(dplyr)

#loop through the file list with lapply
dfs <-lapply(files, function(file) {
   #read file
   page <- read_xml(file)
   
   #get id for each file
   id <- xml_find_first(out, "//ip") %>% xml_attr("id_pac") 
   
   #get information from each the requested nodes
   rodcis <- xml_find_first(out, ".//rodcis") %>% xml_text()
   datod <- xml_find_first(out, ".//dat_od") %>% xml_text()
   
   #make data frame of results for each file
   data.frame(id, rodcis, datod)
})

#combine all results into 1 data frame
answer <- bind_rows(dfs)