SQLAlchemy:一个查询中有多个计数 [英] SQLAlchemy: several counts in one query

问题描述

我很难优化我的SQLAlchemy查询.我的SQL知识非常基础，而且我无法从SQLAlchemy文档中获得所需的东西.

I am having hard time optimizing my SQLAlchemy queries. My SQL knowledge is very basic, and I just can't get the stuff I need from the SQLAlchemy docs.

假设以下非常基本的一对多关系:

Suppose the following very basic one-to-many relationship:

class Parent(Base):
    __tablename__ = "parents"
    id = Column(Integer, primary_key = True)
    children = relationship("Child", backref = "parent")

class Child(Base):
    __tablename__ = "children"
    id = Column(Integer, primary_key = True)
    parent_id = Column(Integer, ForeignKey("parents.id"))
    naughty = Column(Boolean)

我怎么做:

• 查询每个父级的(Parent, count_of_naughty_children, count_of_all_children)元组吗?

• Query tuples of (Parent, count_of_naughty_children, count_of_all_children) for each parent?

在花费大量时间进行谷歌搜索之后，我发现了如何分别查询这些值:

After decent time spent googling, I found how to query those values separately:

# The following returns tuples of (Parent, count_of_all_children):
session.query(Parent, func.count(Child.id)).outerjoin(Child, Parent.children).\
    group_by(Parent.id)
# The following returns tuples of (Parent, count_of_naughty_children):
al = aliased(Children, session.query(Children).filter_by(naughty = True).\
    subquery())
session.query(Parent, func.count(al.id)).outerjoin(al, Parent.children).\
    group_by(Parent.id)

我试图以不同的方式将它们组合在一起，但没有设法得到我想要的东西.

I tried to combine them in different ways, but didn't manage to get what I want.

• 查询所有顽皮孩子超过80％的父母吗?顽皮可能为NULL.

我猜想此查询将基于上一个查询，并按顽皮/所有比率进行过滤.

I guess this query is going to be based on the previous one, filtering by naughty/all ratio.

感谢您的帮助.

:感谢Antti Haapala的帮助，我找到了第二个问题的解决方案:

EDIT : Thanks to Antti Haapala's help, I found solution to the second question:

avg = func.avg(func.coalesce(Child.naughty, 0)) # coalesce() treats NULLs as 0
# avg = func.avg(Child.naughty) - if you want to ignore NULLs
session.query(Parent).join(Child, Parent.children).group_by(Parent).\
    having(avg > 0.8)

它找到孩子的naughty变量的平均值，将False和NULL视为0，将True视为1.在MySQL后端进行了测试，但也应该适用于其他变量.

It finds average if children's naughty variable, treating False and NULLs as 0, and True as 1. Tested with MySQL backend, but should work on others, too.

推荐答案

count() sql aggretate函数非常简单；它为您提供每个组中非空值的总数.考虑到这一点，我们可以调整您的查询以为您提供适当的结果.

the count() sql aggretate function is pretty simple; it gives you the total number of non-null values in each group. With that in mind, we can adjust your query to give you the proper result.

print (Query([
    Parent,
    func.count(Child.id),
    func.count(case(
        [((Child.naughty == True), Child.id)], else_=literal_column("NULL"))).label("naughty")])

    .join(Parent.children).group_by(Parent)
    )

哪个会产生以下sql:

Which produces the following sql:

SELECT 
 parents.id AS parents_id, 
 count(children.id) AS count_1, 
 count(CASE WHEN (children.naughty = 1) 
       THEN children.id 
       ELSE NULL END) AS naughty 
FROM parents 
JOIN children ON parents.id = children.parent_id 
GROUP BY parents.id

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