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lxml xpath syntax to access the ancestor of an XML element of specific depth?

July 31, 2023

I am trying to access ancestors of depth 3 in an XML file, i.e. for element /a/b/c/d/e/f, I want to get element c.

Here is my more realistic example input file:

<?xml version="1.0" encoding="utf-8"?>
<Project xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="urn:QDA-XML:project:1.0">
  <Sources>
    <TextSource name="document example">
      <Description />
      <PlainTextSelection>
        <Description />
        <Coding>
          <CodeRef targetGUID="a2a627dd-f7e7-4fc7-b8db-918e3ad50450" />
        </Coding>
      </PlainTextSelection>
    </TextSource>
    <VideoSource name="myvideo">
      <Transcript>
        <SyncPoint/>
        <SyncPoint/>
        <TranscriptSelection>
          <Description />
          <Coding>
            <CodeRef targetGUID="a2a627dd-f7e7-4fc7-b8db-918e3ad50450" />
          </Coding>
        </TranscriptSelection>
      </Transcript>
      <VideoSelection>
        <Coding>
          <CodeRef targetGUID="a2a627dd-f7e7-4fc7-b8db-918e3ad50450" />
        </Coding>
      </VideoSelection>
    </VideoSource>
  </Sources>
  <Notes>
    <Note name="some text">
      <Description />
      <PlainTextSelection>
        <Description />
        <Coding>
          <CodeRef targetGUID="a2a627dd-f7e7-4fc7-b8db-918e3ad50450" />
        </Coding>
      </PlainTextSelection>
    </Note>
  </Notes>
</Project>

In this case for instance, I want to access the elements Note, TextSource and VideoSource that are ancestors of CodeRef elements.

I have the following working code, but am wondering if there is a nicer way to go about it, perhaps using Xpath syntax:

import lxml.etree as ET

tree = ET.parse('coderef_examples/project_simplified.xml')
root = tree.getroot()

for i in root.findall('.//CodeRef', root.nsmap):
    p = tree.getelementpath(i)
    p = p.replace('{urn:QDA-XML:project:1.0}', '')
    print('namespace-free path: ', p)

    p = tree.getpath(i) # Xpath
    s = '/'.join(p.split('/')[:4]) # Xpath of depth 3
    print('xpath string: ', s)
    ancestor = root.xpath(s)[0]
    print('source tag: ', ancestor.tag, ', source name: ', ancestor.get('name'))

Output:

namespace-free path:  Sources/TextSource/PlainTextSelection/Coding/CodeRef
xpath string:  /*/*[1]/*[1]
source tag:  {urn:QDA-XML:project:1.0}TextSource , source name:  document example
namespace-free path:  Sources/VideoSource/Transcript/TranscriptSelection/Coding/CodeRef
xpath string:  /*/*[1]/*[2]
source tag:  {urn:QDA-XML:project:1.0}VideoSource , source name:  myvideo
namespace-free path:  Sources/VideoSource/VideoSelection/Coding/CodeRef
xpath string:  /*/*[1]/*[2]
source tag:  {urn:QDA-XML:project:1.0}VideoSource , source name:  myvideo
namespace-free path:  Notes/Note/PlainTextSelection/Coding/CodeRef
xpath string:  /*/*[2]/*
source tag:  {urn:QDA-XML:project:1.0}Note , source name:  some text

Can it be done directly in Xpath? (ideally in a way that is independent of the tag of the ancestor and the depth of the CodeRef elements)

edit: Solution based on Conal Tuohy’s answer:

import lxml.etree as ET

tree = ET.parse('coderef_examples/project_simplified.xml')
root = tree.getroot()

for ancestor in root.xpath('/*/*/*[descendant::qda:CodeRef]', namespaces={'qda': 'urn:QDA-XML:project:1.0'}):
    print('source tag: ', ancestor.tag, ', source name: ', ancestor.get('name'))

Much faster and much more efficient. It may not print an entry for every CodeRef node, but since this is what I want in my case, it is even better.