access ElementTree node parent node

I am using the builtin Python ElementTree module. It is straightforward to access children, but what about parent or sibling nodes? - can this be done efficiently without traversing the entire tree?

There's no direct support in the form of a parent attribute, but you can perhaps use the patterns described here to achieve the desired effect. The following one-liner is suggested (updated from the linked-to post to Python 3.8) to create a child-to-parent mapping for a whole tree, using the method xml.etree.ElementTree.Element.iter:

parent_map = {c: p for p in tree.iter() for c in p}

Vinay's answer should still work, but for Python 2.7+ and 3.2+ the following is recommended:

parent_map = {c:p for p in tree.iter() for c in p}

getiterator() is deprecated in favor of iter(), and it's nice to use the new dict list comprehension constructor.

Secondly, while constructing an XML document, it is possible that a child will have multiple parents, although this gets removed once you serialize the document. If that matters, you might try this:

parent_map = {}
for p in tree.iter():
    for c in p:
        if c in parent_map:
            parent_map[c].append(p)
            # Or raise, if you don't want to allow this.
        else:
            parent_map[c] = [p]
            # Or parent_map[c] = p if you don't want to allow this

You can use xpath ... notation in ElementTree.

<parent>
     <child id="123">data1</child>
</parent>

xml.findall('.//child[@id="123"]...')
>> [<Element 'parent'>]

As mentioned in https://stackoverflow.com/questions/24239435/get-parent-element-after-using-find-method-xml-etree-elementtree you would have to do an indirect search for parent. Having xml:

<a>
 <b>
  <c>data</c>
  <d>data</d>    
 </b>
</a>

Assuming you have created etree element into xml variable, you can use:

 In[1] parent = xml.find('.//c/..')
 In[2] child = parent.find('./c')

Resulting in:

Out[1]: <Element 'b' at 0x00XXXXXX> 
Out[2]: <Element 'c' at 0x00XXXXXX>

Higher parent would be found as:secondparent=xml.find('.//c/../..') being <Element 'a' at 0x00XXXXXX>

Pasting here my answer from https://stackoverflow.com/a/54943960/492336:

I had a similar problem and I got a bit creative. Turns out nothing prevents us from adding the parent info ourselves. We can later strip it once we no longer need it.

def addParentInfo(et):
    for child in et:
        child.attrib['__my_parent__'] = et
        addParentInfo(child)

def stripParentInfo(et):
    for child in et:
        child.attrib.pop('__my_parent__', 'None')
        stripParentInfo(child)

def getParent(et):
    if '__my_parent__' in et.attrib:
        return et.attrib['__my_parent__']
    else:
        return None

# Example usage

tree = ...
addParentInfo(tree.getroot())
el = tree.findall(...)[0]
parent = getParent(el)
while parent:
    doSomethingWith(parent)
    parent = getParent(parent)
stripParentInfo(tree.getroot())

The XPath '..' selector cannot be used to retrieve the parent node on 3.5.3 nor 3.6.1 (at least on OSX), eg in interactive mode:

import xml.etree.ElementTree as ET
root = ET.fromstring('<parent><child></child></parent>')
child = root.find('child')
parent = child.find('..') # retrieve the parent
parent is None # unexpected answer True

The last answer breaks all hopes...

Got an answer from

https://towardsdatascience.com/processing-xml-in-python-elementtree-c8992941efd2

Tip: use '...' inside of XPath to return the parent element of the current element.

for object_book in root.findall('.//*[@name="The Hunger Games"]...'):
    print(object_book)

If you are using lxml, I was able to get the parent element with the following:

parent_node = next(child_node.iterancestors())

This will raise a StopIteration exception if the element doesn't have ancestors - so be prepared to catch that if you may run into that scenario.

Another way if just want a single subElement's parent and also known the subElement's xpath.

parentElement = subElement.find(xpath+"/..")

Look at the 19.7.2.2. section: Supported XPath syntax ...

Find node's parent using the path:

parent_node = node.find('..')