All combinations of set of dictionaries into K N-sized groups

This is where I’d leverage the new dataclasses with sets. You can make a dataclass hashable by setting frozen=True in the decorator. First you’d add your players to a set to get unique players. Then you’d get all the combinations of players for n size teams. Then you could create a set of unique teams. Then create valid groupings whereas no player is represented more than once across teams. Finally you could calculate the max disparity in the total team skill level across the grouping (leveraging combinations yet again) and use that to sort your valid groupings. So something like this.

from dataclasses import dataclass
from itertools import combinations
from typing import FrozenSet

import yaml


@dataclass(order=True, frozen=True)
class Player:
    name: str
    skill: int


@dataclass(order=True, frozen=True)
class Team:
    members: FrozenSet[Player]

    def total_skill(self):
        return sum(p.skill for p in self.members)


def is_valid(grouping):
    players = set()
    for team in grouping:
        for player in team.members:
            if player in players:
                return False
            players.add(player)
    return True


def max_team_disparity(grouping):
    return max(
        abs(t1.total_skill() - t2.total_skill())
        for t1, t2 in combinations(grouping, 2)
    )


def best_team_matchups(player_file, k, n):
    with open(player_file) as f:
        players = set(Player(p['name'], p['skill']) for p in yaml.load(f))
    player_combs = combinations(players, n)
    unique_teams = set(Team(frozenset(team)) for team in player_combs)
    valid_groupings = set(g for g in combinations(unique_teams, k) if is_valid(g))
    for g in sorted(valid_groupings, key=max_team_disparity):
        print(g)


best_team_matchups('test.yaml', k=2, n=4)

Example output:

(
    Team(members=frozenset({
        Player(name='Chr', skill=6),
        Player(name='Christopher', skill=6),
        Player(name='Nicholas', skill=7),
        Player(name='Patricia', skill=4)
    })),
    Team(members=frozenset({
        Player(name='Bia', skill=4),
        Player(name='Bianca', skill=4),
        Player(name='Danny', skill=8),
        Player(name='Nicho', skill=7)
    }))
)

Instead of trying to create every possible grouping of k sets of n elements (possibly including repeats!), and then filtering down to the ones that don’t have any overlap, let’s directly build groupings that meet the criterion. This also avoids generating redundant groupings in different orders (the original code could also do this by using combinations rather than product in the last step).

The approach is:

• Iterate over possibilities (combinations of n elements in the input) for the first set – by which I mean, the one that contains the first of the elements that will be chosen.
• For each, recursively find possibilities for the remaining sets. They cannot use elements from the first set, and they also cannot use elements from before the first set (or else the first set wouldn’t be first).

In order to combine the results elegantly, we use a recursive generator: rather than trying to build lists that contain results from the recursive calls, we just yield everything we need to. We represent each collection of group_count many elements with a tuple of tuples (the inner tuples are the groups). At the base case, there is exactly one way to make no groups of elements – by just… doing that… yeah… – so we need to yield one value which is a tuple of no tuples of an irrelevant number of elements each – i.e., an empty tuple. In the other cases, we prepend the tuple for the current group to each result from the recursive call, yielding all those results.

from itertools import combinations

def non_overlapping_groups(group_count, group_size, population):
    if group_count == 0:
        yield ()
        return
    for indices in combinations(range(len(population)), group_size):
        current = (tuple(population[i] for i in indices),)
        remaining = [
            x for i, x in enumerate(population)
            if i not in indices and i > indices[0]
        ] if indices else population
        for recursive in non_overlapping_groups(group_count - 1, group_size, remaining):
            yield current + recursive

Let’s try it:

>>> list(non_overlapping_groups(2, 3, 'abcdef'))
[(('a', 'b', 'c'), ('d', 'e', 'f')), (('a', 'b', 'd'), ('c', 'e', 'f')), (('a', 'b', 'e'), ('c', 'd', 'f')), (('a', 'b', 'f'), ('c', 'd', 'e')), (('a', 'c', 'd'), ('b', 'e', 'f')), (('a', 'c', 'e'), ('b', 'd', 'f')), (('a', 'c', 'f'), ('b', 'd', 'e')), (('a', 'd', 'e'), ('b', 'c', 'f')), (('a', 'd', 'f'), ('b', 'c', 'e')), (('a', 'e', 'f'), ('b', 'c', 'd'))]
>>> list(non_overlapping_groups(3, 2, 'abcdef'))
[(('a', 'b'), ('c', 'd'), ('e', 'f')), (('a', 'b'), ('c', 'e'), ('d', 'f')), (('a', 'b'), ('c', 'f'), ('d', 'e')), (('a', 'c'), ('b', 'd'), ('e', 'f')), (('a', 'c'), ('b', 'e'), ('d', 'f')), (('a', 'c'), ('b', 'f'), ('d', 'e')), (('a', 'd'), ('b', 'c'), ('e', 'f')), (('a', 'd'), ('b', 'e'), ('c', 'f')), (('a', 'd'), ('b', 'f'), ('c', 'e')), (('a', 'e'), ('b', 'c'), ('d', 'f')), (('a', 'e'), ('b', 'd'), ('c', 'f')), (('a', 'e'), ('b', 'f'), ('c', 'd')), (('a', 'f'), ('b', 'c'), ('d', 'e')), (('a', 'f'), ('b', 'd'), ('c', 'e')), (('a', 'f'), ('b', 'e'), ('c', 'd'))]
>>> # Some quick sanity checks
>>> len(list(non_overlapping_groups(2, 3, 'abcdef')))
10
>>> # With fewer input elements, obviously we can't do it.
>>> len(list(non_overlapping_groups(2, 3, 'abcde')))
0
>>> # Adding a 7th element, any element could be the odd one out,
>>> # and in each case we get another 10 possibilities, making 10 * 7 = 70.
>>> len(list(non_overlapping_groups(2, 3, 'abcdefg')))
70

I performance tested this against a modified version of the original (which also shows how to make it work properly with non-strings, and optimizes the sum calculation):

def unique_group(group_count, group_size, population):
    groups = list(it.combinations(population, group_size))
    return (
        i for i in combinations(groups, group_count) 
        if len({e for g in i for e in g}) == group_count * group_size
    )

Quickly verifying the equivalence:

>>> len(list(unique_group(3, 2, 'abcdef')))
15
>>> len(list(non_overlapping_groups(3, 2, 'abcdef')))
15
>>> set(unique_group(3, 2, 'abcdef')) == set(non_overlapping_groups(3, 2, 'abcdef'))
True

We see that even for fairly small examples (here, the output has 280 groupings), the brute-force approach has to filter through a lot:

>>> import timeit
>>> timeit.timeit("list(g(3, 3, 'abcdefghi'))", globals={'g': unique_group}, number=100)
5.895461600041017
>>> timeit.timeit("list(g(3, 3, 'abcdefghi'))", globals={'g': non_overlapping_groups}, number=100)
0.2303082060534507