Question

我正在比赛管理器上，并且开始在比赛系统上工作。我的问题是如何退回两支球队的所有组合，然后才能使用这些组合进行比赛？
让我解释一下。一队仅属于一个池。在“团队”模型中，我有一个外键字段，引用了该团队所属的池。以我的算法，我想创建一个属于同一个池的两支球队参加的比赛。例如，我有A，B，C，D团队，我想创建A对抗B，A对抗C，A对抗D等。

我在这里看了一下：https://docs.python.org/3/library/itertools.html，发现了“组合”功能，仅此而已。

models.py

class Match(models.Model):
    scoreTeam1  = models.IntegerField()
    scoreTeam2  = models.IntegerField()
    phase       = models.ForeignKey(Phase, default=None, on_delete=models.CASCADE)
    teams       = models.ManyToManyField(Team, default=None, blank=True)

class Pool(models.Model):
    name    = models.CharField(max_length=16)
    nbTeam  = models.IntegerField(validators=[
                                    MaxValueValidator(4),
                                    MinValueValidator(3)
                                ])
    phase   = models.ForeignKey(Phase, default=None, on_delete=models.CASCADE)
    field   = models.ForeignKey(Field, default=None, on_delete=models.CASCADE)

class Team(models.Model):
    name        = models.CharField(max_length=16)
    totalpoints = models.IntegerField(default=0)
    position    = models.IntegerField(default=0)
    category    = models.ForeignKey(Category, default=None, on_delete=models.CASCADE)
    pool        = models.ForeignKey(Pool, default=None, blank=True, null=True, on_delete=models.SET_NULL)

views.py

def matches_phase_view(request, id, id_phase, *args, **kwargs):
    tournaments = Tournament.objects.filter(user=request.user)
    tournament = Tournament.objects.get(pk=id)
    phase = Phase.objects.get(pk=id_phase)
    pools = Pool.objects.filter(phase=phase)
    teams = Team.objects.filter(pool__in=pools)

    matches_count = 0
    for pool in pools:
        matches_count = matches_count + pool.nbTeam * (pool.nbTeam - 1) // 2
    for pool in pools:
        teams_combinations = combinations(Team.objects.filter(pool=pool), 2)
        list_teams_combinations = list(teams_combinations)
        for combination in list_teams_combinations:
            match = Match.objects.create(phase=phase)
            match.teams.add(combination[0])
            match.teams.add(combination[1])
            match.save()

    matches = Match.objects.filter(phase=phase)
    context = {
        'tournaments': tournaments,
        'tournament': tournament,
        'pools': pools,
        'teams': teams,
        'matches_count': matches_count,
        'matches': matches,
    }
    return render(request, 'tournament_manage_phase_matches.html', context)

我尝试过：

for pool in pools:
        a = combinations(Team.objects.filter(pool=pool), 2)

但是我遇到了这个错误：

未定义名称“组合”

如果有人有更好的主意，我想听听。

我正在使用Python 3.7，Django 2.2。

编辑：

template

{% extends 'base_tournament.html' %}

{% load crispy_forms_tags %}

{% block title %}
  Tournament
{% endblock %}

{% block navbar_manage %}
    {% include 'navbar_manage.html' %}
{% endblock %}

{% block content %}
    <h1>Matches of ...</h1>

    <ul class="nav nav-tabs" id="myTab" role="tablist">
    {% for pool in pools %}
        <li class="nav-item">
            <a class="nav-link" id="{{ pool.name }}-tab" data-toggle="tab" href="#{{ pool.name }}" role="tab" aria-controls="{{ pool.name }}" aria-selected="true">{{ pool.name }}</a>
        </li>
    {% endfor %}
    </ul>
    <div class="tab-content" id="myTabContent">
        {% for pool in pools %}
            <div class="tab-pane fade" id="{{ pool.name }}" role="tabpanel" aria-labelledby="{{ pool.name }}-tab">
                {% for match in matches %}
                    <p>Wait for a moment</p>
                {% endfor %}
            </div>
        {% endfor %}
    </div>

{% endblock %}

Answer 1

from itertools import combinations

如何在Django中返回两个团队的所有组合？

1 个答案: