Something like this?

>>> from xml.sax.saxutils import escape
>>> escape("< & >")   
'&lt; &amp; &gt;'

One thing that I tried, that worked for me is to open the xml file as a file object , then use ElementTree.fromstring() passing in the complete contents of the file.

Example -

>>> import xml.etree.ElementTree as ET
>>> ef = ET.parse('a.xml')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "C:\Python34\lib\xml\etree\ElementTree.py", line 1187, in parse
    tree.parse(source, parser)
  File "C:\Python34\lib\xml\etree\ElementTree.py", line 598, in parse
    self._root = parser._parse_whole(source)
ValueError: multi-byte encodings are not supported
>>> with open('a.xml','r') as f:
...     ef = ET.fromstring(f.read())
...
>>> ef
<Element 'productMeta' at 0x028DF180>

You can also, create an XMLParser with the required encoding, and this should enable you to be able to parse strings from that encoding, Example -

import xml.etree.ElementTree as ET
xmlp = ET.XMLParser(encoding="utf-8")
f = ET.parse('a.xml',parser=xmlp)

Im pretty sure there is no builtins that accomplish exactly what you ask

I think the best you could do is just iterate over the characters and "fix" each one (see example that i think is complete)

try: # python2
  from htmlentitydefs import codepoint2name
except: # python3
  from html.entities import codepoint2name

def encode_xml(c):
  # return the character or its &#XX; or &entity; representation
  ascii_val = ord(c)
  known_entity =  codepoint2name.get(ascii_val,None)
  if known_entity: # this is a named codepoint
    return "&%s;"%(known_entity,)  
  # printable characters are ascii values [32..127] inclusive
  is_normal_character =  32 <= ascii_val <= 127
  if is_normal_character:
      return c
  return hex(ascii_val).replace("0x","&#")+";"


def make_xml_entity_string(s):
  return "".join(encode_xml(c) for c in s)

print("R:", make_xml_entity_string( 'abc < ABC\x14\xF2&'))

you could then go the otherway ... in roughly the same manner (taking advantage of regex this time though)

try: # python2  
  from htmlentitydefs import name2codepoint
except: # python3
  from html.entities import name2codepoint
import re

def decode_xml_replacer(match):
  name=match.group(1);
  if(name.startswith("#")):
    return chr(int(name[1:],16))
  return chr(name2codepoint.get(name,'?'))

def decode_xml_string(s):
  return re.sub("&(.*?);",decode_xml_replacer,s)

... note that this wont work for codepoints > 255 i think

>>> from xml.dom.minidom import Document
>>> a=Document()
>>> a.toprettyxml(encoding="utf-8")
'<?xml version="1.0" encoding="utf-8"?>\n'

or

>>> a.toxml(encoding="utf-8")
'<?xml version="1.0" encoding="utf-8"?>'

you can set the encoding for the document.writexml() function in the same way.

data is a unicode str. The error is saying that such a thing which also contains an encoding="..." declaration is not supported, because a str is supposedly already decoded from its encoding and hence it's ambiguous/nonsensical that it would also contain an encoding declaration. It's telling you to use a bytes instead, e.g. data = b'<...>'. Presumably you should be opening a file in binary mode, read the data from there and let etree handle the encoding="...", instead of using string literals in your code, which complicates the encoding situation even further.

It's as simple as:

from xml.etree import ElementTree

#        open in binary mode ↓
with open('/tmp/test.xml', 'rb') as f:
    e = ElementTree.fromstring(f.read())

Et voilà, e contains your parsed file with the encoding having been (presumably) correctly interpreted by etree based on the file's internal encoding="..." header.

ElementTree in fact has a shortcut method for this:

e = ElementTree.parse('/tmp/test.xml')

The standard lib’s very own HTMLParser has an undocumented function unescape() which does exactly what you think it does:

up to Python 3.4:

import HTMLParser
h = HTMLParser.HTMLParser()
h.unescape('© 2010') # u'\xa9 2010'
h.unescape('© 2010') # u'\xa9 2010'

Python 3.4+:

import html
html.unescape('© 2010') # u'\xa9 2010'
html.unescape('© 2010') # u'\xa9 2010'