Hey Ben,

look at my post above. I use this approach at the moment and it's suitable. To fit the rank best, even if an item is not ranked within RTC, do a first sorting by the work item id.

I give you a short Python code snippet I use at the moment for the ranking. Maybe it's useful.

First, I put all items (in fact I use the history items, but that's not important) into a list and calculate a 4-tuple for each item like this:

aptPriority_nodes = history_node.findall("allExtensions[key='com.ibm.team.apt.attribute.planitem.priority._pm7NmRYUEd6L1tNIGdz5qQ']/smallStringValue")
        priority = history_node.find('priority/id').text
        id_rank_string = "%012d"%int(item.identifier)
        if len(aptPriority_nodes) == 0 or aptPriority_nodes[0].text == None:
            aptPriority = None
            aptRank = None
        else:
            apt_parts = aptPriority_nodes[0].text.partition(' ')
            aptPriority = apt_parts[0]
            aptRank = apt_parts[2]
        rank = (priority,aptPriority,aptRank,id_rank_string)

Than I can get a suitable ranking with this sorting:

        for item in itemlist:

            if item.get_rank()[2] == None:
                logging.warn("Item %(id)s should get ranked but no ranking is provided by RTC! Ranking may be not in sync with RTC for this item."%({'id':item.identifier}))
            if item.get_rank()[1] != None and item.get_rank()[1] != item.get_rank()[0]:
                logging.warn("The priority of item %(id)s was set back to  “Unassigned” that leads to an RTC internal inconsistency. Ranking may be not in sync with RTC for this item."%({'id':item.identifier}))
        # first oder: by id, descending
        itemlist.sort(key=lambda item: item.get_rank()[3],reverse=True)
        # second order: by rank string
        itemlist.sort(key=lambda item: item.get_rank()[2])
        # by ranked or unranked
        itemlist.sort(key=lambda item: item.get_rank()[1],reverse=True)
        # last order: by priority literal, descending
        itemlist.sort(key=lambda item: item.get_rank()[0],reverse=True)

It works for me. It's not so hard to investigate the logic. It's sad that RTC has no official and documented way to do that. I think it's a core functionality of real Backlog Management Tools.

Hope that helps.