Python: ElementTree, get the namespace string of an Element

Question

This XML file is named example.xml:

<?xml version="1.0"?>
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/maven-v4_0_0.xsd">

  <modelVersion>14.0.0</modelVersion>
  <groupId>.com.foobar.flubber</groupId>
  <artifactId>uberportalconf</artifactId>
  <version>13-SNAPSHOT</version>
  <packaging>pom</packaging>
  <name>Environment for UberPortalConf</name>
  <description>This is the description</description>    
  <properties>
      <birduberportal.version>11</birduberportal.version>
      <promotiondevice.version>9</promotiondevice.version>
      <foobarportal.version>6</foobarportal.version>
      <eventuberdevice.version>2</eventuberdevice.version>
  </properties>
  <!-- A lot more here, but as it is irrelevant for the problem I have removed it -->
</project>

If I load example.xml and parse it with ElementTree I can see its namespace is http://maven.apache.org/POM/4.0.0.

>>> from xml.etree import ElementTree
>>> tree = ElementTree.parse('example.xml')
>>> print tree.getroot()
<Element '{http://maven.apache.org/POM/4.0.0}project' at 0x26ee0f0>

I have not found a method to call to get just the namespace from an Element without resorting to parsing the str(an_element) of an Element. It seems like there got to be a better way.

do you know how to use the find method in this cases? it didnt work here... — caarlos0
– caarlos0, Commented May 5, 2012 at 3:48

Mark Ransom · Accepted Answer · 2022-07-27 12:40:35Z

35

This is a perfect task for a regular expression.

import re

def namespace(element):
    m = re.match(r'\{.*\}', element.tag)
    return m.group(0) if m else ''

edited Jul 27, 2022 at 12:40

answered Oct 18, 2012 at 3:34

Mark Ransom

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7 Comments

Robert Over a year ago

After fighting for a while with this issue, this is the best solution I found. I can't believe that the API don't get you a way to ask for the namespace and, at the same time, it doesn't return the attribute 'xmlns' when doing 'rootElement.keys()'. Sure there is a good reason for that but I can't find it at this moment.

JustWe Over a year ago

add r before regular exp please make this answer perfect.

Mark Ransom Over a year ago

@Jiu thank you so much. I can't believe I missed that.

Piotr Boho Over a year ago

to get namespace without curly braces included: re.match(r'\{(.*)\}', element.tag).group(1)

FNia Over a year ago

After adding r before the regex the \s are redundant.

|

dimakin · Accepted Answer · 2024-03-24 09:35:17Z

32

The namespace should be in Element.tag right before the "actual" tag:

>>> root = tree.getroot()
>>> root.tag
'{http://maven.apache.org/POM/4.0.0}project'

To know more about namespaces, take a look at ElementTree: Working with Namespaces and Qualified Names.

edited Mar 24, 2024 at 9:35

dimakin

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answered Mar 1, 2012 at 9:44

Rik Poggi

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Comments

Jakub Roztocil · Accepted Answer · 2012-03-01 09:43:01Z

13

I am not sure if this is possible with xml.etree, but here is how you could do it with lxml.etree:

>>> from lxml import etree
>>> tree = etree.parse('example.xml')
>>> tree.xpath('namespace-uri(.)')
'http://maven.apache.org/POM/4.0.0'

answered Mar 1, 2012 at 9:43

Jakub Roztocil

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7 Comments

Deleted Over a year ago

I get unresolved import: etree using Python 2.7.2 in Windows. xpath wasn´t available as a method when using xml.etree and if I use find() (which supports xpath expressions) the 'namespace-uri(.)' statement still doesn´t work.

Andreas Scherer Over a year ago

this is exactly what i was looking for, see pr on gh

kieran Over a year ago

This has been the best solution that I've seen. I normally use xmlstarlet but I may switch now.

ccpizza Over a year ago

for lxml a simpler way to get the namespace is tree.getroot().nsmap

ccpizza Over a year ago

@Jona: I'd assume that using None is a way to address the default namespace, ie the one which is declared without a prefix.

|

Lorcan · Accepted Answer · 2015-02-02 17:29:09Z

11

Without using regular expressions:

>>> root
<Element '{http://www.google.com/schemas/sitemap/0.84}urlset' at 0x2f7cc10>

>>> root.tag.split('}')[0].strip('{')
'http://www.google.com/schemas/sitemap/0.84'

edited Feb 2, 2015 at 17:29

answered Feb 2, 2015 at 17:17

Lorcan

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1 Comment

watashiSHUN Over a year ago

a similar answer root.tag[1:root.tag.index('}')]

Cypher · Accepted Answer · 2016-08-30 02:27:17Z

2

The lxml.xtree library's element has a dictionary called nsmap, which shows all the namespace that are in use in the current tag scope.

>>> item = tree.getroot().iter().next()
>>> item.nsmap
{'md': 'urn:oasis:names:tc:SAML:2.0:metadata'}

answered Aug 30, 2016 at 2:27

Cypher

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Comments

Community · Accepted Answer · 2017-05-23 11:47:01Z

The short answer is:

ElementTree._namspace_map[ElementTree._namspace_map.values().index('')]

but only if you have been calling

ElementTree.register_namespace(prefix,uri)

in response to every event=="start-ns" received while iterating through the result of

ET.iterparse(...)

and you registered for "start-ns"

The answer the question "what is the default namespace?", it is necessary to clarify two points:

(1) XML specifications say that the default namespace is not necessarily global throughout the tree, rather the default namespace can be re-declared at any element under root, and inherits downwards until meeting another default namespace re-declaration.

(2) The ElementTree module can (de facto) handle XML-like documents which have no root default namespace, -if- they have no namespace use anywhere in the document. (* there may be less strict conditions, e.g., that is "if" and not necessarily "iff").

It's probably also worth considering "what do you want it for?" Consider that XML files can be semantically equivalent, but syntactically very different. E.g., the following three files are semantically equivalent, but A.xml has one default namespace declaration, B.xml has three, and C.xml has none.

A.xml:
<a xlmns="http://A" xlmns:nsB0="http://B0" xlmns:nsB1="http://B1">
     <nsB0:b/>
     <nsB1:b/>
</a>

B.xml:
<a xlmns="http://A">
     <b xlmns="http://B0"/>
     <b xlmns="http://B1"/>
</a>

C.xml:
<{http://A}a>
     <{http://B0}b/>
     <{http://B1}b/>
</a>

The file C.xml is the canonical expanded syntactical representation presented to the ElementTree search functions.

If you are certain a priori that there will be no namespace collisions, you can modify the element tags while parsing as discussed here: Python ElementTree module: How to ignore the namespace of XML files to locate matching element when using the method "find", "findall"

jcollado · Accepted Answer · 2012-03-01 09:44:04Z

1

I think it will be easier to take a look at the attributes:

>>> root.attrib
{'{http://www.w3.org/2001/XMLSchema-instance}schemaLocation':
   'http://maven.apache.org/POM/4.0.0 http://maven.apache.org/maven-v4_0_0.xsd'}

answered Mar 1, 2012 at 9:44

jcollado

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2 Comments

Deleted Over a year ago

Certainly easier than parsing str(the_element). But I guess parsing the_element.tag is even a bit easier. As I am only interested in the namespace. What do you think?

jcollado Over a year ago

I think that @RikPoggi's answer seems the best one (actually, I upvoted it). In fact, getting the namespace should be as easy as re.search('\{(.*)\}', the_element.tag).group(1). With my answer it looks you could use the_element.attrib.values()[0].split()[0], but, indeed, it doesn't look so much straightforward and it isn't guaranteed that you won't get any other attributes in the future.

rez · Accepted Answer · 2021-05-21 11:25:27Z

0

Here is my solution on ElementTree 3.9+,

def get_element_namespaces(filename, element):
    namespace = []
    for key, value in ET.iterparse(filename, events=['start', 'start-ns']):
        print(key, value)
        if key == 'start-ns':
            namespace.append(value)
        else:
            if ET.tostring(element) == ET.tostring(value):
                return namespace
            namespace = []
    return namespaces

This would return an array of [prefix:URL] tuples like this:

[('android', 'http://schemas.android.com/apk/res/android'), ('tools', 'http://schemas.android.com/tools')]

answered May 21, 2021 at 11:25

rez

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1 Comment

not2qubit Over a year ago

This can't run because you have typos and no context in there.

user8874906 · Accepted Answer · 2017-11-02 13:21:05Z

-1

combining some of the answers above, I think the shortest code is

theroot = tree.getroot()
theroot.attrib[theroot.keys()[0]]

answered Nov 2, 2017 at 13:21

user8874906

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1 Comment

Mitch Lindgren Over a year ago

This is not accurate, because the xmlns might not be the first attribute of the root. In fact, I'm currently trying to parse a TCX file and the xmlns isn't showing up as an attribute of the root at all.

Comprised · Accepted Answer · 2025-11-13 18:07:58Z

-1

This is how you can get all root level namespaces. Credit goes to Google's AI Overview

import xml.etree.ElementTree as ET

xml_data = """<?xml version="1.0"?>
<root xmlns:ns1="http://example.com/ns1" xmlns:ns2="http://example.com/ns2">
    <ns1:element1>Text 1</ns1:element1>
    <ns2:element2>Text 2</ns2:element2>
</root>
"""

# Parse the XML string incrementally and capture namespace declarations
namespaces = dict([node for _, node in ET.iterparse(xml_data, events=['start-ns'])])

# Print the collected namespaces
for prefix, uri in namespaces.items():
    print(f"Prefix: '{prefix}', URI: '{uri}'")```

answered Nov 13 at 18:07

Comprised

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