Alter namespace prefixing with ElementTree in Python

Question

By default, when you call ElementTree.parse(someXMLfile) the Python ElementTree library prefixes every parsed node with it\'s namespace URI in Clark\'s Notation:

Answer 1

xml.etree.ElementTree doesn't appear to have fixtag, well, not according to the documentation. However I've looked at some source code for fixtag and you do:

import xml.etree.ElementTree as ET

for event, elem in ET.iterparse(inFile, events=("start", "end")):
    namespace, looktag = string.split(elem.tag[1:], "}", 1)

You have the tag string in looktag, suitable for a lookup. The namespace is in namespace.

Answer 2

You don't specifically need to use iterparse. Instead, the following script:

from cStringIO import StringIO
import xml.etree.ElementTree as ET

NS_MAP = {
    'http://www.red-dove.com/ns/abc' : 'rdc',
    'http://www.adobe.com/2006/mxml' : 'mx',
    'http://www.red-dove.com/ns/def' : 'oth',
}

DATA = '''<?xml version="1.0" encoding="utf-8"?>
<rdc:container xmlns:mx="http://www.adobe.com/2006/mxml"
                 xmlns:rdc="http://www.red-dove.com/ns/abc"
                 xmlns:oth="http://www.red-dove.com/ns/def">
  <mx:Style>
    <oth:style1/>
  </mx:Style>
  <mx:Style>
    <oth:style2/>
  </mx:Style>
  <mx:Style>
    <oth:style3/>
  </mx:Style>
</rdc:container>'''

tree = ET.parse(StringIO(DATA))
some_node = tree.getroot().getchildren()[1]
print ET.fixtag(some_node.tag, NS_MAP)
some_node = some_node.getchildren()[0]
print ET.fixtag(some_node.tag, NS_MAP)

produces

('mx:Style', None)
('oth:style2', None)

Which shows how you can access the fully-qualified tag names of individual nodes in a parsed tree. You should be able to adapt this to your specific needs.