Setting and accessing namespaces in python lxml

Question

I am writing a script that processes a rdf:skos file with python3 and lxml:

I learnt that I need to pass to the findall procedure the namespaces that the XML mentions. (Ok, strange, since the XML files lists these in the header, so this seems like an unnecessary step but anyway).

When calling

for concept in root.findall('.//skos:Concept', namespaces=root.nsmap):

that works, because a root.nsmap is constructed by lxml.

But then later in my code I also need to perform a test on xml:lang

for pl in concept.findall(".//skos:prefLabel[@xml:lang='en']", namespaces=root.nsmap):

and here python tells me

SyntaxError: prefix 'xml' not found in prefix map

Ok, true, in my skos file there is no extra declaration for the xml namespace. So I try to add it to the root.nsmap dict

root.nsmap['xml'] = "http://www.w3.org/XML/1998/namespace"

but that too doesn't work

nsmap = {'rdf': 'http://www.w3.org/1999/02/22-rdf-syntax-ns#', 'uneskos': 'http://purl.org/umu/uneskos#', 'iso-thes': 'http://purl.org/iso25964/skos-thes#', 'dcterms': 'http://purl.org/dc/terms/', 'skos': 'http://www.w3.org/2004/02/skos/core#', 'rdfs': 'http://www.w3.org/2000/01/rdf-schema#'}

Seems I am not allowed to modify the root.nsmap?

Anyone an idea how this is done? I have processed tons of XML in the past with Perl XML::Twig which is very very comfortable and I assmue, the Python community has (at least) similarly comfortable ways to do that ... but how?

Any hint appreciated.

mzjn · Accepted Answer · 2018-01-06 17:00:33Z

1

Modifying root.nsmap has no effect. But you can create another dictionary and modify that one. Example:

from lxml import etree

doc = """
<root xmlns:skos="http://www.w3.org/2004/02/skos/core#">
   <skos:prefLabel xml:lang='en'>FOO</skos:prefLabel>
   <skos:prefLabel xml:lang='de'>BAR</skos:prefLabel>
</root>"""

root = etree.fromstring(doc)
nsmap = root.nsmap
nsmap["xml"] = "http://www.w3.org/XML/1998/namespace" 

en = root.find(".//skos:prefLabel[@xml:lang='en']", namespaces=nsmap)
print(en.text)

Output:

FOO

answered Jan 6, 2018 at 17:00

mzjn

51.5k16 gold badges140 silver badges266 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

user3310299 Over a year ago

Thanks a lot! works like a breeze. Issue solved.(Still, if I may, I find it strange that lxml a) forces me to make special arrangements for the default namespace "xml" that the original XML file does not even declare and b) that I can't extend the nsmap that the parser creates ...) Anyway, glad it works. Thanks again!!

mzjn Over a year ago

I agree that it is strange that you are forced to declare the "xml" prefix in the Python code when it does not need to be declared in the XML file.

Collectives™ on Stack Overflow

Setting and accessing namespaces in python lxml

1 Answer 1

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related