Compare XML snippets?

You can use formencode.doctest_xml_compare -- the xml_compare function compares two ElementTree or lxml trees.

The order of the elements can be significant in XML, this may be why most other methods suggested will compare unequal if the order is different... even if the elements have same attributes and text content.

But I also wanted an order-insensitive comparison, so I came up with this:

from lxml import etree
import xmltodict  # pip install xmltodict


def normalise_dict(d):
    """
    Recursively convert dict-like object (eg OrderedDict) into plain dict.
    Sorts list values.
    """
    out = {}
    for k, v in dict(d).iteritems():
        if hasattr(v, 'iteritems'):
            out[k] = normalise_dict(v)
        elif isinstance(v, list):
            out[k] = []
            for item in sorted(v):
                if hasattr(item, 'iteritems'):
                    out[k].append(normalise_dict(item))
                else:
                    out[k].append(item)
        else:
            out[k] = v
    return out


def xml_compare(a, b):
    """
    Compares two XML documents (as string or etree)

    Does not care about element order
    """
    if not isinstance(a, basestring):
        a = etree.tostring(a)
    if not isinstance(b, basestring):
        b = etree.tostring(b)
    a = normalise_dict(xmltodict.parse(a))
    b = normalise_dict(xmltodict.parse(b))
    return a == b