在Django中按属性组合两个模型的最快方法

Question

我有两个具有共同属性的模型，如果两个属性相等，我想两个组合它们，每个模型从不同的数据库中获取数据：

第一个模型

class User(models.Model):
      matricula = models.CharField(max_length=100)

      def __unicode__(self):
          return self.matricula

第二模型

class UserMoodle(models.Model):
      matricula = models.CharField(max_length=100)

      def __unicode__(self):
          return self.matricula

实施例

如果我有：

lists_users = [<User: 123>, <User: 2345>,<User:567>]
lists_users_moodle =  [<UserMoodle:123>, <UserMoodle:2345>, <UserMoodle:897>]

我想将它们组合起来得到这个结果：

combined_models_lists = [[<User: 123>, <UserMoodle: 123>], [<User: 2345>, <UserMoodle: 2345>], [<User: 567>, None],[None, <UserMoodle: 897>]]

提前致谢！

Answer 1

一种方法是首先从这些列表（或QuerySet）创建dicts，然后迭代这些dicts的键的并集以获得所需的输出：

dict_users = {x.matricula : x for x in lists_users}
dict_moodle = {x.matricula : x for x in lists_users_moodle}

combined_models_lists  = [[dict_users.get(x), dict_moodle.get(x)] 
                                          for x in set(dict_users).union(dict_moodle)]

---

另一种方法是迭代每个QuerySet的排序版本，然后对列表进行填充，这样我们将使用比上述方法更少的内存：

users = User.objects.all().order_by('matricula') #order by field `matricula`
moodles = UserMoodle.objects.all().order_by('matricula')

out = []

it1 = iter(users)
it2 = iter(moodles)

prev1 = next(it1)
prev2 = next(it2)

while True:
    if prev1.matricula  == prev2.matricula:
        out.append([prev1, prev2])
        try:
            prev1 = next(it1)
            prev2 = next(it2)
        except StopIteration:
            out.extend([x, None] for x in it1)
            out.extend([None, x] for x in it2)
            break

    if prev1.matricula  < prev2.matricula :
        out.append([prev1, None])
        try:
            prev1 = next(it1)
        except StopIteration:
            out.append([None, prev2])
            out.extend([None, x] for x in it2)
            break

    if prev1.matricula  > prev2.matricula :
        out.append([None, prev2])
        try:
            prev2 = next(it2)
        except StopIteration:
            out.extend([prev1, None])
            out.extend([x, None] for x in it1)
            break