Convert xml without namespaces

Question

Thanks to Dave2e's answer (Convert XML to dataframe) I managed to convert most of my xml-files to a dataframe. However, I have several files, that don't use namespaces. The script doesn't show any error messages, but it doesn't run through either (after 48 hours I cancelled the process). The files are about 58Mb (each).

How do I have to change the code so that it runs through?

  doc <- read_xml('/path/to/Quelle/Response.xml')
  
  # namespace identifizieren
  ns <- xml_ns(doc)
  
  reportedPersons <- xml_find_all(doc, ".//d1:reportedPerson")
  
  dfs<- lapply(reportedPersons, function(reportedPerson) {
    
    #find all the children nodes
    allchildren <- reportedPerson %>% xml_find_all( './/*') 
    
    #count the number of children in each node
    branches <-reportedPerson %>% xml_find_all( './/*') %>% xml_length(only_elements = TRUE)
    
    #filter to keep the lowest level - nodes with values - no children
    leafs <- allchildren[which(branches==0)]
    
    #get values and names of the leafs
    values <- leafs %>% xml_text()
    names <- leafs %>% xml_name() #xml_name(ns=ns)
    #get names of the parents - still results in a few duplicates
    parentNames_top <-  sapply(leafs, function(leaf {xml_parent(xml_parent(xml_parent(xml_parent(leaf)))) %>% xml_name()})
    parentNames_middle1 <-  sapply(leafs, function(leaf){xml_parent(xml_parent(xml_parent(leaf))) %>% xml_name()})
    parentNames_middle2 <-  sapply(leafs, function(leaf){xml_parent(xml_parent(leaf)) %>% xml_name()})
    parentNames <-  sapply(leafs, function(leaf){xml_parent(leaf) %>% xml_name()})
    x <- paste(parentNames_top, parentNames_middle1, parentNames_middle2, parentNames, names, sep="_")
    
    #put into a dataframe 
    df <- as.data.frame(t(values), cut.names = FALSE)
    names(df) = paste(parentNames_top, parentNames, names, sep="_")
    #there are still a few duplicates could use full path names to avoid duplicates
    #names(df) = xml_path(leafs)
    
    df
  })
  
  d.test <- bind_rows(dfs)

Excerp from a file:



  
    1-2k78-1
    3-LI-1
    2-TZ-3
    53453-FGHRFTZH
    1
    
      In AG
      tres
      2020
    
    
      45der4
      1
      1
    
    2021-04-12T11:31:37.52
    2021-03-31
    2021-04-12
    1
    false
  
  
    
      
        
          6664442863124
          
            WW.454
            3335
          
          
            RT.5674
            99999999
          
          Test
          Wels
          9
          
            1733-01-01
          
        
        
          Test
          Wels
          Welsi
        
        
          
            1733-01-01
          
          
            
              
                4353
                YX
                Yukiland
              
              Eikwrl
            
          
          9
        
        
          000
          1733-01-01
        
        
          9
        
        
          9
          
            
              9655
              ZT
              Zentralris
            
            1733-01-01
          
        
        de
        
          Goswiesl
          ZT
          893
        
      
      
        
          
            3453
            Twers
            ZT
            93434
          
          1733-01-01
          
            
              45335
              Wekwr
              EO
              234923
            
          
          
            99999999
            999
            99999
            
              Werkhof
              1
              Tewrw
              5656
              00
              23423
              UZ
            
            1
            1733-01-01

Parfait · Accepted Answer

Consider XSLT (special-purpose language designed to transform XML files) to flatten your XML at node across all levels of its descendants. To avoid repeated name, the grandparent and parent node names are concatenated to current node name. Below works on this XML and your previous posted XML. R can run XSLT 1.0 with the xslt package (sister package to xml2):

XSLT (save as .xsl file, a special .xml file)

Online Demo

R

library(xml2)
library(xslt)

# LOAD XML AND XSL
doc <- read_xml("/path/to/Input.xml")
style <- read_xml("/path/to/Flatten/Script.xsl", package = "xslt")

# RUN TRANSFORMATION
new_xml <- xml_xslt(doc, style)

# RETRIEVE reportedPerson NODES
nmsp <- c(doc = "http://www.ech.ch/xmlns/eCH-0099/2")
recs <- xml2::xml_find_all(new_xml, "//doc:reportedPerson", ns=nmsp)

# BIND EACH CHILD TEXT AND NAME
df_list <- lapply(recs, function(r) {
  vals <- xml2::xml_children(r)
  
  data.frame(rbind(setNames(c(xml2::xml_text(vals)), 
                            c(xml2::xml_name(vals)))))
})

# COMBINE ALL DFS
reportedPerson_df <- do.call(rbind.data.frame, df_list)

# REMOVE HELPER OBJECTS
rm(recs, df_list)

Alternatively, if you have your own XSLT processor and can run raw XML through it or call with R's system() at command line, the original XML package can work as well with handy xmlToDataFrame. This demonstrates the portability of XSLT!

library(XML)

# COMMAND LINE CALL TO UNIX'S XSLTPROC (ALTERNATIVE TO xslt PACKAGE)
system("xsltproc -o /path/to/input.xml /path/to/xslt_script.xsl /path/to/output.xml")

doc <- xmlParse("/path/to/output.xml")
nmsp <- c(doc = "http://www.ech.ch/xmlns/eCH-0099/2")

reportedPerson_df <- xmlToDataFrame(
    doc, nodes=getNodeSet(doc, "//doc:reportedPerson", namespace=nmsp)
)

Convert xml without namespaces

Answers (1)

Related Questions