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copied!<p>Have you tried sorting the list first? The algorithm you proposed should work, albeit with lots of database hits.</p> <pre><code>import itertools cars = [ {'car': 'X2', 'mods': [1,2]}, {'car': 'Y2', 'mods': [2]}, {'car': 'W2', 'mods': [1]}, {'car': 'X1', 'mods': [1,2]}, {'car': 'W1', 'mods': [1]}, {'car': 'Y1', 'mods': [2]}, {'car': 'Z1', 'mods': [1,2,3]}, {'car': 'X3', 'mods': [1,2]}, ] cars.sort(key=lambda car: car['mods']) cars_by_common_mods = {} for k, g in itertools.groupby(cars, lambda car: car['mods']): cars_by_common_mods[frozenset(k)] = [car['car'] for car in g] print cars_by_common_mods </code></pre> <p>Now, about those queries:</p> <pre><code>import collections import itertools from operator import itemgetter from django.db import connection cursor = connection.cursor() cursor.execute('SELECT car_id, mod_id FROM someapp_car_mod ORDER BY 1, 2') cars = collections.defaultdict(list) for row in cursor.fetchall(): cars[row[0]].append(row[1]) # Here's one I prepared earlier, which emulates the sample data we've been working # with so far, but using the car id instead of the previous string. cars = { 1: [1,2], 2: [2], 3: [1], 4: [1,2], 5: [1], 6: [2], 7: [1,2,3], 8: [1,2], } sorted_cars = sorted(cars.iteritems(), key=itemgetter(1)) cars_by_common_mods = [] for k, g in itertools.groupby(sorted_cars, key=itemgetter(1)): cars_by_common_mods.append({'mods': k, 'cars': map(itemgetter(0), g)}) print cars_by_common_mods # Which, for the sample data gives me (reformatted by hand for clarity) [{'cars': [3, 5], 'mods': [1]}, {'cars': [1, 4, 8], 'mods': [1, 2]}, {'cars': [7], 'mods': [1, 2, 3]}, {'cars': [2, 6], 'mods': [2]}] </code></pre> <p>Now that you've got your lists of car ids and mod ids, if you need the complete objects to work with, you could do a single query for each to get a complete list for each model and create a lookup <code>dict</code> for those, keyed by their ids - then, I believe, Bob is your proverbial father's brother.</p>

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