Access ElementTree node parent node

Question

I am using the Python ElementTree built-in module. Direct access to children, but how to relate to parent or sister nodes? - can this be done efficiently without going through the whole tree?

Answer 1

There is no direct support in the form of the parent attribute, but you can use the templates described here to achieve the desired effect. The following one-line (from a related message) is proposed to create a child-parent mapping for the entire tree:

 parent_map = dict((c, p) for p in tree.getiterator() for c in p) 

Answer 2

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

Secondly, when creating an XML document, it is possible that the child will have several parents, although this will be deleted after the document is serialized. If that matters, you can try the following:

 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 

Answer 3

You can use xpath ... notation in ElementTree.

 <parent> <child id="123">data1</child> </parent> xml.findall('.//child[@id="123"]...') >> [<Element 'parent'>] 

Answer 4

As mentioned in Get the parent after using the find method (xml.etree.ElementTree) , you have to do an indirect search for the parent. Having xml:

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

Assuming you created an etree element in an xml variable, you can use:

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

Result:

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

A higher parent will be found as: secondparent=xml.find('.//c/../..') <Element 'a' at 0x00XXXXXX>

Answer 5

Another way, if only one parent subElement is required, and the path to the subElement subelement is also known.

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

Answer 6

If you use lxml, I managed to get the parent element with the following:

 parent_node = next(child_node.iterancestors()) 

Answer 7

Look at clause 19.7.2.2. section: Supported XPath Syntax ...

Find the parent node using the path:

 parent_node = node.find('..')