问题描述
我需要计算每个卖家 ID 的期间中位数(参见下面的简化模型).问题是我无法构造 ORM 查询.
I need to calculate period medians per seller ID (see simplyfied model below). The problem is I am unable to construct the ORM query.
型号
class MyModel:
period = models.IntegerField(null=True, default=None)
seller_ids = ArrayField(models.IntegerField(), default=list)
aux = JSONField(default=dict)
查询
queryset = (
MyModel.objects.filter(period=25)
.annotate(seller_id=Func(F("seller_ids"), function="unnest"))
.values("seller_id")
.annotate(
duration=Cast(KeyTextTransform("duration", "aux"), IntegerField()),
median=Func(
F("duration"),
function="percentile_cont",
template="%(function)s(0.5) WITHIN GROUP (ORDER BY %(expressions)s)",
),
)
.values("median", "seller_id")
)
ArrayField 聚合(seller_id
)来源
我认为我需要做的是以下几行
I think what I need to do is something along the lines below
select t.*, p_25, p_75
from t join
(select district,
percentile_cont(0.25) within group (order by sales) as p_25,
percentile_cont(0.75) within group (order by sales) as p_75
from t
group by district
) td
on t.district = td.district
以上示例来源
Python 3.7.5、Django 2.2.8、Postgres 11.1
Python 3.7.5, Django 2.2.8, Postgres 11.1
推荐答案
您可以像 Ryan Murphy (https://gist.github.com/rdmurphy/3f73c7b1826cacee34f6c2a855b12e2e).Median
然后就像 Avg
:
You can create a Median
child class of the Aggregate
class as was done by Ryan Murphy (https://gist.github.com/rdmurphy/3f73c7b1826cacee34f6c2a855b12e2e). Median
then works just like Avg
:
from django.db.models import Aggregate, FloatField
class Median(Aggregate):
function = 'PERCENTILE_CONT'
name = 'median'
output_field = FloatField()
template = '%(function)s(0.5) WITHIN GROUP (ORDER BY %(expressions)s)'
然后找到一个字段使用的中位数
Then to find the median of a field use
my_model_aggregate = MyModel.objects.all().aggregate(Median('period'))
然后以 my_model_aggregate['period__median']
的形式提供.
which is then available as my_model_aggregate['period__median']
.
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