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Python Exclusive XML Canonicalization (xml-exc-c14n)

In Python, I need to Canonicalize (c14n) an XML string.
Which module/package can I use for this? And how should I do this?
(I prefer to use default python 2.7 modules, without extra installs or patches.)
For reference see: http://www.w3.org/TR/xml-exc-c14n/

from http://www.decalage.info/en/python/lxml-c14n
lxml provides a very easy way to do c14n in python.
<..>
Here is an example showing how to perform C14N using lxml 2.1:
import lxml.etree as ET
et = ET.parse('file.xml')
output = StringIO.StringIO()
et.write_c14n(output)
print output.getvalue()
from lxml docs:
write_c14n(self, file, exclusive=False, with_comments=True,
compression=0, inclusive_ns_prefixes=None)
C14N write of document. Always writes UTF-8.
<..>
Also there is libxml2:
XML C14N version 1.0 provides two options which make four
possibilities (see http://www.w3.org/TR/xml-c14n and
http://www.w3.org/TR/xml-exc-c14n/):
Inclusive or Exclusive C14N
With or without comments
libxml2 gives access to these options in its C14N API:
http://xmlsoft.org/html/libxml-c14n.html
Though obligatory check for version changes in these two libs.

now in python 3 you can write your code like this:
import lxml.etree as ET
et = ET.parse('your_xml_file_that_you_want_to_canonicalize.xml')
et.write_c14n("your_result_will_be_in_this_file.xml")

element tree treats similar files differently

Here are two different files that my python (2.6) script encounters. One will parse, the other will not. I'm just curious as to why this happens.
This xml file will not parse and the script will fail:
<Landfire_Feedback_Point_xlsform id="fbfm40v10" instanceID="uuid:9e062da6-b97b-4d40-b354-6eadf18a98ab" submissionDate="2013-04-30T23:03:32.881Z" isComplete="true" markedAsCompleteDate="2013-04-30T23:03:32.881Z" xmlns="http://opendatakit.org/submissions">
<date_test>2013-04-17</date_test>
<plot_number>10</plot_number>
<select_multiple_names>BillyBob</select_multiple_names>
<geopoint_plot>43.2452830500 -118.2149402900 210.3000030518 3.0000000000</geopoint_plot><fbfm40_new>GS2</fbfm40_new>
<select_grazing>NONE</select_grazing>
<image_close>1366230030355.jpg</image_close>
<plot_note>No road present.</plot_note>
<n0:meta xmlns:n0="http://openrosa.org/xforms">
<n0:instanceID>uuid:9e062da6-b97b-4d40-b354-6eadf18a98ab</n0:instanceID>
</n0:meta>
</Landfire_Feedback_Point_xlsform>
This xml file will parse correctly and the script succeeds:
<Landfire_Feedback_Point_xlsform id="fbfm40v10">
<date_test>2013-05-14</date_test>
<plot_number>010</plot_number>
<select_multiple_names>BillyBob</select_multiple_names>
<geopoint_plot>43.26630563 -118.39881809 351.70001220703125 5.0</geopoint_plot>
<fbfm40_new>GR1</fbfm40_new>
<select_grazing>HIGH</select_grazing>
<image_close>fbfm40v10_PLOT_010_ID_6.jpg</image_close>
<plot_note>Heavy grazing</plot_note>
<meta><instanceID>uuid:90e7d603-86c0-46fc-808f-ea0baabdc082</instanceID></meta>
</Landfire_Feedback_Point_xlsform>
Here is a little python script that demonstrates that one will work, while the other will not. I'm just looking for an explanation as to why one is seen by ElementTree as an xml file while the other isn't. Specifically, the one that doesn't seem to parse fails with a "'NONE' type doesn't have a 'text' attribute" or something similar. But, it's because it doesn't seem to consider the file as xml or it can't see any elements beyond the opening line. Any explanation or direction with regard to this error would be appreciated. Thanks in advance.
Python script:
import os
from xml.etree import ElementTree
def replace_xml_attribute_in_file(original_file,element_name,attribute_value):
#THIS FUNCTION ONLY WORKS ON XML FILES WITH UNIQUE ELEMENT NAMES
# -DUPLICATE ELEMENT NAMES WILL ONLY GET THE FIRST ELEMENT WITH A GIVEN NAME
#split original filename and add tempfile name
tempfilename="temp.xml"
rootsplit = original_file.rsplit('\\') #split the root directory on the backslash
rootjoin = '\\'.join(rootsplit[:-1]) #rejoin the root diretory parts with a backslash -minus the last
temp_file = os.path.join(rootjoin,tempfilename)
et = ElementTree.parse(original_file)
author=et.find(element_name)
author.text = attribute_value
et.write(temp_file)
if os.path.exists(temp_file) and os.path.exists(original_file): #if both the original and the temp files exist
os.remove(original_file) #erase the original
os.rename(temp_file,original_file) #rename the new file
else:
print "Something went wrong."
replace_xml_attribute_in_file("testfile1.xml","image_close","whoopdeedoo.jpg");

Here is a little python script that demonstrates that one will work, while the other will not. I'm just looking for an explanation as to why one is seen by ElementTree as an xml file while the other isn't.
Your code doesn't demonstrate that at all. It demonstrates that they're both seen by ElementTree as valid XML files chock full of nodes. They both parse just fine, they both read past the first line, etc.
The only problem is that the first one doesn't have a node named 'image_close', so your code doesn't work.
You can see that pretty easily:
for node in et.getroot().getchildren():
print node.tag
You get 9 children of the root, with either version.
And the output to that should show you the problem. The node you want is actually named {http://opendatakit.org/submissions}image_close in the first example, rather than image_close as in the second.
And, as you can probably guess, this is because of the namespace=http://opendatakit.org/submissions in the root node. ElementTree uses the "James Clark notation" for mapping unknown-namespaced names to universal names.
Anyway, because none of the nodes are named image_close, the et.find(element_name) returns None, so your code stores author=None, then tries to assign to author.text, and gets an error.
As for how to fix this problem… well, you could learn how namespaces work by default in ElementTree, or you could upgrade to Python 2.7 or install a newer ElementTree for 2.6 that lets you customize things more easily. But if you want to do custom namespace handling and also stick with your old version… I'd start with this article (and its two predecessors) and this one.

Use Python to edit XML header

I've written a Python script to create some XML, but I didn't find a way to edit the heading within Python.
Basically, instead of having this as my heading:
<?xml version="1.0" ?>
I need to have this:
<?xml version="1.0" encoding="UTF-8" standalone="no" ?>
I looked into this as best I could, but found no way to add standalone status from within Python. So I figured that I'd have to go over the file after it had been created and then replace the text. I read in several places that I should stay far away from using readlines() because it could ruin the XML formatting.
Currently the code I have to do this - which I got from another Stackoverflow post about editing XML with Python - is:
doc = parse('file.xml')
elem = doc.find('xml version="1.0"')
elem.text = 'xml version="1.0" encoding="UTF-8" standalone="no"'
That provides me with a KeyError. I've tried to debug it to no avail, which leads me to believe that perhaps the XML heading wasn't meant to be edited in this way. Or my code is just wrong.
If anyone is curious (or miraculously knows how to insert standalone status into Python code), here is my code to write the xml file:
with open('file.xml', 'w') as f:
f.write(doc.toprettyxml(indent=' '))
Some people don't like "toprettyxml", but with my relatively basic level, it seemed like the best bet.
Anyway, if anyone can provide some advice or insight, I would be very grateful.

The xml.etree API does not give you any options to write out a standalone attribute in the XML declaration.
You may want to use the lxml library instead; it uses the same ElementTree API, but offers more control over the output. tostring() supports a standalone flag:
from lxml import etree
etree.tostring(doc, pretty_print=True, standalone=False)
or use .write(), which support the same options:
doc.write(outputfile, pretty_print=True, standalone=False)

Python ElementTree module: How to ignore the namespace of XML files to locate matching element when using the method "find", "findall"

I want to use the method of findall to locate some elements of the source xml file in the ElementTree module.
However, the source xml file (test.xml) has namespaces. I truncate part of xml file as sample:
<?xml version="1.0" encoding="iso-8859-1"?>
<XML_HEADER xmlns="http://www.test.com">
<TYPE>Updates</TYPE>
<DATE>9/26/2012 10:30:34 AM</DATE>
<COPYRIGHT_NOTICE>All Rights Reserved.</COPYRIGHT_NOTICE>
<LICENSE>newlicense.htm</LICENSE>
<DEAL_LEVEL>
<PAID_OFF>N</PAID_OFF>
</DEAL_LEVEL>
</XML_HEADER>
The sample python code is below:
from xml.etree import ElementTree as ET
tree = ET.parse(r"test.xml")
el1 = tree.findall("DEAL_LEVEL/PAID_OFF") # Return None
el2 = tree.findall("{http://www.test.com}DEAL_LEVEL/{http://www.test.com}PAID_OFF") # Return <Element '{http://www.test.com}DEAL_LEVEL/PAID_OFF' at 0xb78b90>
Though using "{http://www.test.com}" works, it's very inconvenient to add a namespace in front of each tag.
How can I ignore the namespace when using functions like find, findall, ...?

Instead of modifying the XML document itself, it's best to parse it and then modify the tags in the result. This way you can handle multiple namespaces and namespace aliases:
from io import StringIO # for Python 2 import from StringIO instead
import xml.etree.ElementTree as ET
# instead of ET.fromstring(xml)
it = ET.iterparse(StringIO(xml))
for _, el in it:
_, _, el.tag = el.tag.rpartition('}') # strip ns
root = it.root
This is based on the discussion here.

If you remove the xmlns attribute from the xml before parsing it then there won't be a namespace prepended to each tag in the tree.
import re
xmlstring = re.sub(' xmlns="[^"]+"', '', xmlstring, count=1)

The answers so far explicitely put the namespace value in the script. For a more generic solution, I would rather extract the namespace from the xml:
import re
def get_namespace(element):
m = re.match('\{.*\}', element.tag)
return m.group(0) if m else ''
And use it in find method:
namespace = get_namespace(tree.getroot())
print tree.find('./{0}parent/{0}version'.format(namespace)).text

Here's an extension to #nonagon answer (which removes namespace from tags) to also remove namespace from attributes:
import io
import xml.etree.ElementTree as ET
# instead of ET.fromstring(xml)
it = ET.iterparse(io.StringIO(xml))
for _, el in it:
if '}' in el.tag:
el.tag = el.tag.split('}', 1)[1] # strip all namespaces
for at in list(el.attrib.keys()): # strip namespaces of attributes too
if '}' in at:
newat = at.split('}', 1)[1]
el.attrib[newat] = el.attrib[at]
del el.attrib[at]
root = it.root
Obviously this is a permanent defacing of the XML but if that's acceptable because there are no non-unique tag names and because you won't be writing the file needing the original namespaces then this can make accessing it a lot easier

Improving on the answer by ericspod:
Instead of changing the parse mode globally we can wrap this in an object supporting the with construct.
from xml.parsers import expat
class DisableXmlNamespaces:
def __enter__(self):
self.old_parser_create = expat.ParserCreate
expat.ParserCreate = lambda encoding, sep: self.old_parser_create(encoding, None)
def __exit__(self, type, value, traceback):
expat.ParserCreate = self.oldcreate
This can then be used as follows
import xml.etree.ElementTree as ET
with DisableXmlNamespaces():
tree = ET.parse("test.xml")
The beauty of this way is that it does not change any behaviour for unrelated code outside the with block. I ended up creating this after getting errors in unrelated libraries after using the version by ericspod which also happened to use expat.

You can use the elegant string formatting construct as well:
ns='http://www.test.com'
el2 = tree.findall("{%s}DEAL_LEVEL/{%s}PAID_OFF" %(ns,ns))
or, if you're sure that PAID_OFF only appears in one level in tree:
el2 = tree.findall(".//{%s}PAID_OFF" % ns)

In python 3.5 , you can pass the namespace as an argument in find().
For example ,
ns= {'xml_test':'http://www.test.com'}
tree = ET.parse(r"test.xml")
el1 = tree.findall("xml_test:DEAL_LEVEL/xml_test:PAID_OFF",ns)
Documentation link :- https://docs.python.org/3.5/library/xml.etree.elementtree.html#parsing-xml-with-namespaces

I might be late for this but I dont think re.sub is a good solution.
However the rewrite xml.parsers.expat does not work for Python 3.x versions,
The main culprit is the xml/etree/ElementTree.py see bottom of the source code
# Import the C accelerators
try:
# Element is going to be shadowed by the C implementation. We need to keep
# the Python version of it accessible for some "creative" by external code
# (see tests)
_Element_Py = Element
# Element, SubElement, ParseError, TreeBuilder, XMLParser
from _elementtree import *
except ImportError:
pass
Which is kinda sad.
The solution is to get rid of it first.
import _elementtree
try:
del _elementtree.XMLParser
except AttributeError:
# in case deleted twice
pass
else:
from xml.parsers import expat # NOQA: F811
oldcreate = expat.ParserCreate
expat.ParserCreate = lambda encoding, sep: oldcreate(encoding, None)
Tested on Python 3.6.
Try try statement is useful in case somewhere in your code you reload or import a module twice you get some strange errors like
maximum recursion depth exceeded
AttributeError: XMLParser
btw damn the etree source code looks really messy.

If you're using ElementTree and not cElementTree you can force Expat to ignore namespace processing by replacing ParserCreate():
from xml.parsers import expat
oldcreate = expat.ParserCreate
expat.ParserCreate = lambda encoding, sep: oldcreate(encoding, None)
ElementTree tries to use Expat by calling ParserCreate() but provides no option to not provide a namespace separator string, the above code will cause it to be ignore but be warned this could break other things.

Let's combine nonagon's answer with mzjn's answer to a related question:
def parse_xml(xml_path: Path) -> Tuple[ET.Element, Dict[str, str]]:
xml_iter = ET.iterparse(xml_path, events=["start-ns"])
xml_namespaces = dict(prefix_namespace_pair for _, prefix_namespace_pair in xml_iter)
return xml_iter.root, xml_namespaces
Using this function we:
Create an iterator to get both namespaces and a parsed tree object.
Iterate over the created iterator to get the namespaces dict that we can
later pass in each find() or findall() call as sugested by
iMom0.
Return the parsed tree's root element object and namespaces.
I think this is the best approach all around as there's no manipulation either of a source XML or resulting parsed xml.etree.ElementTree output whatsoever involved.
I'd like also to credit balmy's answer with providing an essential piece of this puzzle (that you can get the parsed root from the iterator). Until that I actually traversed XML tree twice in my application (once to get namespaces, second for a root).

Just by chance dropped into the answer here: XSD conditional type assignment default type confusion?. This is not the exact answer for the topic question but may be applicable if the namespace is not critical.
<?xml version="1.0" encoding="UTF-8"?>
<persons xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:noNamespaceSchemaLocation="test.xsd">
<person version="1">
<firstname>toto</firstname>
<lastname>tutu</lastname>
</person>
</persons>
Also see: https://www.w3.org/TR/xmlschema-1/#xsi_schemaLocation
Works for me. I call an XML validation procedure in my application. But also I want to quickly see the validation highliting and autocompletion in PyCharm when editing the XML. This noNamespaceSchemaLocation attribute does what I need.
RECHECKED
from xml.etree import ElementTree as ET
tree = ET.parse("test.xml")
el1 = tree.findall("person/firstname")
print(el1[0].text)
el2 = tree.find("person/lastname")
print(el2.text)
Returnrs
>python test.py
toto
tutu

xml to Python data structure using lxml

How can I convert xml to Python data structure using lxml?
I have searched high and low but can't find anything.
Input example
<ApplicationPack>
<name>Mozilla Firefox</name>
<shortname>firefox</shortname>
<description>Leading Open Source internet browser.</description>
<version>3.6.3-1</version>
<license name="Firefox EULA">http://www.mozilla.com/en-US/legal/eula/firefox-en.html</license>
<ms-license>False</ms-license>
<vendor>Mozilla Foundation</vendor>
<homepage>http://www.mozilla.org/firefox</homepage>
<icon>resources/firefox.png</icon>
<download>http://download.mozilla.org/?product=firefox-3.6.3&os=win&lang=en-GB</download>
<crack required="0"/>
<install>scripts/install.sh</install>
<postinstall name="Clean Up"></postinstall>
<run>C:\\Program Files\\Mozilla Firefox\\firefox.exe</run>
<uninstall>c:\\Program Files\\Mozilla Firefox\\uninstall\\helper.exe /S</uninstall>
<requires name="autohotkey" />
</ApplicationPack>

>>> from lxml import etree
>>> treetop = etree.fromstring(anxmlstring)
converts the xml in the string to a Python data structure, and so does
>>> othertree = etree.parse(somexmlurl)
where somexmlurl is the path to a local XML file or the URL of an XML file on the web.
The Python data structure these functions provide (known as an "element tree", whence the etree module name) is well documented here -- all the classes, functions, methods, etc, that the Python data structure in question supports. It closely matches one supported in the Python standard library, by the way.
If you want some different Python data structure, you'll have to walk through the Python data structure which lxml returns, as above mentioned, and build your different data structure yourself based on the information thus collected; lxml can't specifically help you, except by offering several helpers for finding information in the parsed structure it returns, so that collecting said info is a flexible, easy task (again, see the documentation URL above).

It's not entirely clear what kind of data structure you're looking for, but here's a link to a code sample to convert XML to python dictionary of lists via lxml.etree.