Parsing XML with namespace in Python via 'ElementTree'


I have the following XML which I want to parse using Python’s ElementTree:

<rdf:RDF xml_base=""

    <owl:Class rdf_about="">
        <rdfs:label xml_lang="en">basketball league</rdfs:label>
        <rdfs:comment xml_lang="en">
          a group of sports teams that compete against each other
          in Basketball


I want to find all owl:Class tags and then extract the value of all rdfs:label instances inside them. I am using the following code:

tree = ET.parse("filename")
root = tree.getroot()

Because of the namespace, I am getting the following error.

SyntaxError: prefix 'owl' not found in prefix map

I tried reading the document at but I am still not able to get this working since the above XML has multiple nested namespaces.

Kindly let me know how to change the code to find all the owl:Class tags.

Asked By: Sudar



You need to give the .find(), findall() and iterfind() methods an explicit namespace dictionary:

namespaces = {'owl': ''} # add more as needed

root.findall('owl:Class', namespaces)

Prefixes are only looked up in the namespaces parameter you pass in. This means you can use any namespace prefix you like; the API splits off the owl: part, looks up the corresponding namespace URL in the namespaces dictionary, then changes the search to look for the XPath expression {}Class instead. You can use the same syntax yourself too of course:


Also see the Parsing XML with Namespaces section of the ElementTree documentation.

If you can switch to the lxml library things are better; that library supports the same ElementTree API, but collects namespaces for you in .nsmap attribute on elements and generally has superior namespaces support.

Answered By: Martijn Pieters

Here’s how to do this with lxml without having to hard-code the namespaces or scan the text for them (as Martijn Pieters mentions):

from lxml import etree
tree = etree.parse("filename")
root = tree.getroot()
root.findall('owl:Class', root.nsmap)


5 years later I’m still running into variations of this issue. lxml helps as I showed above, but not in every case. The commenters may have a valid point regarding this technique when it comes merging documents, but I think most people are having difficulty simply searching documents.

Here’s another case and how I handled it:

<?xml version="1.0" ?><Tag1 >

… <rdfs:label xml_lang=”en”>basketball league</rdfs:label>
… <rdfs:comment xml_lang=”en”>
… a group of sports teams that compete against each other
… in Basketball
… </rdfs:comment>
… </owl:Class>

… </rdf:RDF>”’
>>> my_namespaces = dict([
… node for _, node in ElementTree.iterparse(
… StringIO(my_schema), events=[‘start-ns’]
… )
… ])
>>> from pprint import pprint
>>> pprint(my_namespaces)
{”: ‘’,
‘owl’: ‘’,
‘rdf’: ‘’,
‘rdfs’: ‘’,
‘xsd’: ‘’}

Then the dictionary can be passed as argument to the search functions:

root.findall('owl:Class', my_namespaces)
Answered By: Brad Dre

I’ve been using similar code to this and have found it’s always worth reading the documentation… as usual!

findall() will only find elements which are direct children of the current tag. So, not really ALL.

It might be worth your while trying to get your code working with the following, especially if you’re dealing with big and complex xml files so that that sub-sub-elements (etc.) are also included.
If you know yourself where elements are in your xml, then I suppose it’ll be fine! Just thought this was worth remembering.


“Element.findall() finds only elements with a tag which are direct children of the current element. Element.find() finds the first child with a particular tag, and Element.text accesses the element’s text content. Element.get() accesses the element’s attributes:”

Answered By: MJM

To get the namespace in its namespace format, e.g. {myNameSpace}, you can do the following:

root = tree.getroot()
ns = re.match(r'{.*}', root.tag).group(0)

This way, you can use it later on in your code to find nodes, e.g using string interpolation (Python 3).

link = root.find(f"{ns}link")
Answered By: Bram Vanroy

My solution is based on @Martijn Pieters’ comment:

register_namespace only influences serialisation, not search.

So the trick here is to use different dictionaries for serialization and for searching.

namespaces = {
    '': '',
    'spec': '',

Now, register all namespaces for parsing and writing:

for name, value in namespaces.iteritems():
    ET.register_namespace(name, value)

For searching (find(), findall(), iterfind()) we need a non-empty prefix. Pass these functions a modified dictionary (here I modify the original dictionary, but this must be made only after the namespaces are registered).

self.namespaces['default'] = self.namespaces['']

Now, the functions from the find() family can be used with the default prefix:

print root.find('default:myelem', namespaces)



does not use any prefixes for elements in the default namespace.

Answered By: peter.slizik

This is basically Davide Brunato’s answer however I found out that his answer had serious problems the default namespace being the empty string, at least on my python 3.6 installation. The function I distilled from his code and that worked for me is the following:

from io import StringIO
from xml.etree import ElementTree
def get_namespaces(xml_string):
    namespaces = dict([
            node for _, node in ElementTree.iterparse(
                StringIO(xml_string), events=['start-ns']
    namespaces["ns0"] = namespaces[""]
    return namespaces

where ns0 is just a placeholder for the empty namespace and you can replace it by any random string you like.

If I then do:

my_namespaces = get_namespaces(my_schema)
root.findall('ns0:SomeTagWithDefaultNamespace', my_namespaces)

It also produces the correct answer for tags using the default namespace as well.

Answered By: M.D.