django - 使用 Django ORM (CROSS JOIN) 计算组合

Question

我有三个相关的模型:Process , Factor和 Level .一个 Process与 Factor 具有多对多关系s 和 Factor将有一个或多个 Level s。我正在尝试计算 Level 的所有组合s 涉及 Process .这很容易用 Python 的 itertools 实现。作为模型方法，但执行速度有点慢，所以我想弄清楚如何使用 Django ORM 在 SQL 中执行此计算。
楷模:

class Process(models.Model):
    factors = models.ManyToManyField(Factor, blank = True)

class Factor(models.Model):
    ...

class Level(models.Model):
    factor = models.ForeignKey(Factor, on_delete=models.CASCADE)

示例:进程 'Running'涉及三个 Factor s ( 'Distance' , 'Climb' , 'Surface' ) 每个由多个 Level 组成s ( 'Long'/ 'Short' , 'Flat'/ 'Hilly' , 'Road'/ 'Mixed'/ 'Trail' )。计算 SQL 中的组合将涉及通过首先确定多少 Factor 来构建查询。 s 参与(在本例中为 3)并执行 CROSS JOIN所有级别的多次。
在 SQL 中，这可以这样完成:

WITH foo AS
    (SELECT * FROM Level
     WHERE Level.factor_id IN
        (SELECT ProcessFactors.factor_id FROM ProcessFactors WHERE process_id = 1)
    )
SELECT a1.*, a2.*, a3.*
    FROM foo a1
    CROSS JOIN foo a2
    CROSS JOIN foo a3
WHERE (a1.factor_id < a2.factor_id) AND (a2.factor_id < a3.factor_id)

结果:

a1.name | a2.name | a3.name
--------------------------
Long    | Flat    | Road
Long    | Flat    | Mixed
Long    | Flat    | Trail
Long    | Hilly   | Road
Long    | Hilly   | Mixed
Long    | Hilly   | Trail
Short   | Flat    | Road
Short   | Flat    | Mixed
Short   | Flat    | Trail
Short   | Hilly   | Road
Short   | Hilly   | Mixed
Short   | Hilly   | Trail

目前，我已将此实现为 Process 上的一种方法。模型为:

def level_combinations(self):
    levels = []
    for factor in self.factors.all():
        levels.append(Level.objects.filter(factor = factor))
    
    combinations = []
    for levels in itertools.product(*levels):
        combination = {}
        
        combination["levels"] = levels
        
        combinations.append(combination)
    
    return combinations

这是否可以使用 Django ORM 或者它是否足够复杂以至于它应该作为原始查询实现以提高 Python 代码实现的速度？
有一个关于 performing CROSS JOIN in Django ORM 的类似问题从几年前(大约 Django v1.3 看起来)似乎没有引起太多关注(作者决定只使用 Python itertools)。

Answer 1

from itertools import groupby, product
    
def level_combinations(self):
    # We need order by factor_id for proper grouping
    levels = Level.objects.filter(factor__process=self).order_by('factor_id')
    # [{'name': 'Long', 'factor_id': 1, ...},
    #  {'name': 'Short', 'factor_id': 1, ...},
    #  {'name': 'Flat', 'factor_id': 2, ...},
    #  {'name': 'Hilly', 'factor_id': 2, ...}]

    groups = [list(group) for _, group in groupby(levels, lambda l: l.factor_id)]
    # [[{'name': 'Long', 'factor_id': 1, ...},
    #   {'name': 'Short', 'factor_id': 1, ...}],
    #  [{'name': 'Flat', 'factor_id': 2, ...},
    #   {'name': 'Hilly', 'factor_id': 2, ...}]]

    # Note: don't forget, that product is iterator/generator, not list
    return product(*groups)

如果顺序无关紧要，那么:

def level_combinations(self):
    levels = Level.objects.filter(factor__process=self)
    groups = {}
    for level in levels:
        groups.setdefault(level.factor_id, []).append(level)
    return product(*groups.values())