Question

This is a part algorithm-logic question (how to do it), part implementation question (how to do it best!). I’m working with Django, so I thought I’d share with that.

In Python, it’s worth mentioning that the problem is somewhat related to how-do-i-use-pythons-itertoolsgroupby.

Suppose you’re given two Django Model-derived classes:

from django.db import models  class Car(models.Model):     mods = models.ManyToManyField(Representative) 

and

from django.db import models  class Mods(models.Model):    ... 

How does one get a list of Cars, grouped by Cars with a common set of Mods?

I.e. I want to get a class likeso:

Cars_by_common_mods = [    { mods: { 'a' }, cars: { 'W1', 'W2' } },   { mods: { 'a', 'b' }, cars: { 'X1', 'X2', 'X3' }, },   { mods: { 'b' }, cars: { 'Y1', 'Y2' } },   { mods: { 'a', 'b', 'c' }, cars: { 'Z1' } }, ] 

I’ve been thinking of something like:

def cars_by_common_mods():   cars = Cars.objects.all()    mod_list = []          for car in cars:     mod_list.append( { 'car': car, 'mods': list(car.mods.all()) }     ret = []    for key, mods_group in groupby(list(mods), lambda x: set(x.mods)):     ret.append(mods_group)    return ret 

However, that doesn’t work because (perhaps among other reasons) the groupby doesn’t seem to group by the mods sets. I guess the mod_list has to be sorted to work with groupby. All to say, I’m confident there’s something simple and elegant out there that will be both enlightening and illuminating.

Cheers & thanks!

Answer 1

Have you tried sorting the list first? The algorithm you proposed should work, albeit with lots of database hits.

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 

Now, about those queries:

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]}] 

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 dict for those, keyed by their ids – then, I believe, Bob is your proverbial father’s brother.