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Home/ Questions/Q 9301023
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Editorial Team
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Editorial Team
Asked: June 18, 20262026-06-18T22:55:13+00:00 2026-06-18T22:55:13+00:00

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

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

<rdf:RDF xml:base="http://dbpedia.org/ontology/"
    xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
    xmlns:owl="http://www.w3.org/2002/07/owl#"
    xmlns:xsd="http://www.w3.org/2001/XMLSchema#"
    xmlns:rdfs="http://www.w3.org/2000/01/rdf-schema#"
    xmlns="http://dbpedia.org/ontology/">

    <owl:Class rdf:about="http://dbpedia.org/ontology/BasketballLeague">
        <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>

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

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

SyntaxError: prefix 'owl' not found in prefix map

I tried reading the document at http://effbot.org/zone/element-namespaces.htm 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.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-18T22:55:16+00:00Added an answer on June 18, 2026 at 10:55 pm

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

    namespaces = {'owl': 'http://www.w3.org/2002/07/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 {http://www.w3.org/2002/07/owl}Class instead. You can use the same syntax yourself too of course:

    root.findall('{http://www.w3.org/2002/07/owl#}Class')
    

    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.

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