Crack a distinguished name string in Python

crackdn.py

def crackdn(dn_str):
    """Crack a DN string into a dict

    'C=US, ST=New Jersey, L=Newark, O=The Dump, Inc., CN=www.thedump.newark'

    ... Becomes ...

    {
      "C": "US",
      "ST": "New Jersey",
      "L": "Newark",
      "O": "The Dump, Inc.",
      "CN": "www.thedump.newark"
    }

    This is much, much simpler and cleaner if you don't need to handle components
    that have commass- e.g. if "The Dump, Inc" was "The Dump Inc." this could
    be done with an easily readable one-liner. Extra logic is required to account
    for commas that may be inside fields
    """
    if not dn_str:
        return None
    # This logic is required to normalize the DN string because
    # we aren't given quoted or escaped values so they may contain
    # commas that will break the basic cracking logic
    # There is probably a cleaner way to do this but this is
    # clear and works reliably
    dn_list = []
    for elt in dn_str.split(','):
        if '=' in elt:
            dn_list.append(elt)
            continue
        assert dn_list

        dn_list.append(','.join([dn_list.pop(), elt]))
    # The simple part
    return dict([kvstr.strip().split('=') for kvstr in dn_list])