Compare XML snippets?

Question

Building on another SO question, how can one check whether two well-formed XML snippets are semantically equal. All I need is \"equal\" or not, since I\'m using this for un

Answer 1

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