我有三个相关的模型:流程,因素和水平。流程与因子有多对多关系,因子有一个或多个级别。我试图计算与流程相关的所有级别组合。使用Python的itertools作为模型方法可以直接实现,但执行速度有点慢,所以我试图弄清楚如何使用Django ORM在SQL中执行此计算。
型号:
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:进程正在运行涉及三个因素(距离,爬升,表面),每个因素由一个级别数量(长/短,平/丘陵,道路/混合/路径)。在SQL中计算组合将涉及通过首先确定涉及多少因子(本例中为3)并多次执行所有级别的CROSS JOIN
来构建查询。
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)
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,我将此作为Process模型的方法实现为:
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或者是否足够复杂以至于应该将其作为原始查询实现以提高Python代码实现的速度?
几年前有一个关于performing CROSS JOIN
in Django ORM的类似问题(看起来大概是Django v1.3)并没有引起吸引人的注意(作者只是使用Python itertools)
答案 0 :(得分:2)
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)
如果订单无关紧要,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())
答案 1 :(得分:1)
如果我理解正确,你可以尝试:
for process in Process.objects.all():
# get all levels for current process
levels = Level.objects.filter(factor__in=process.factors.all())
答案 2 :(得分:0)
几年后,这种解决方法没有实际使用CROSS JOIN
,但它确实在单个中产生了所需的结果em> 查询。
第 1 步:将 cross
字段添加到您的 Factor
模型
class Factor(models.Model):
cross = models.ForeignKey(
to='self', on_delete=models.CASCADE, null=True, blank=True)
...
第 2 步:使用新的 'Climb'
字段将 'Surface'
链接到 'Distance'
,并将 'Climb'
链接到 Factor.cross
第三步:查询如下
Level.objects.filter(factor__name='Distance').values_list(
'name', 'factor__cross__level__name', 'factor__cross__cross__level__name')
结果:
('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')
这是一个简化的例子。为了使其更通用,您可以添加一个带有两个外键的新 Factor.cross
模型,而不是添加 CrossedFactors
字段到 Factor
。然后可以使用该模型来定义各种实验设计。