Question

这是我的SQLAlchemy查询代码

medium_contact_id_subq = (g.session.query(distinct(func.unnest(FUContact.medium_contact_id_lis))).filter(FUContact._id.in_(contact_id_lis))).subquery()
q = (g.session.query(FUMessage).
         filter(FUMessage.fu_medium_contact_id.in_(medium_contact_id_subq))
         .order_by(desc(FUMessage.timestamp_utc))
         )

我想将FUMessage的{{1}}限制为medium_contact_id N个结果。

作为一种解决方法，这是我目前丑陋且未经优化的代码：

    medium_contact_id_lis = (g.session.query(distinct(func.unnest(FUContact.medium_contact_id_lis))).filter(FUContact._id.in_(contact_id_lis))).all()
    q = None
    for medium_contact_id_tup in medium_contact_id_lis:
        medium_contact_id = medium_contact_id_tup[0]
        if q is None:
            q = (g.session.query(FUMessage)
                 .filter(FUMessage.fu_medium_contact_id == medium_contact_id)
                 .limit(MESSAGE_LIMIT)
                 )
        else:
            subq = (g.session.query(FUMessage)
                 .filter(FUMessage.fu_medium_contact_id == medium_contact_id)
                 .limit(MESSAGE_LIMIT)
                 )
            q = q.union(subq)
    q = q.order_by(desc(FUMessage.timestamp_utc))

Answer 1

在每个组中获取热门 N 行的一种方法是在子选择中使用rank()或row_number()等窗口函数，并使用所需的分组和顺序，然后按在封闭选择中。对于 N = 1 ，您可以在Postgresql中使用DISTINCT ON ... ORDER BY组合。

使用函数元素的over()方法生成窗口表达式可以直接使用SQLAlchemy：

medium_contact_id_subq = g.session.query(
        func.unnest(FUContact.medium_contact_id_lis).distinct()).\
    filter(FUContact._id.in_(contact_id_lis)).\
    subquery()

# Perform required filtering in the subquery. Choose a suitable ordering,
# or you'll get indeterminate results.
subq = g.session.query(
        FUMessage,
        func.row_number().over(
            partition_by=FUMessage.fu_medium_contact_id,
            order_by=FUMessage.timestamp_utc).label('n')).\
    filter(FUMessage.fu_medium_contact_id.in_(medium_contact_id_subq)).\
    subquery()

fumessage_alias = aliased(FUMessage, subq)

# row_number() counts up from 1, so include rows with a row num
# less than or equal to limit
q = g.session.query(fumessage_alias).\
    filter(subq.c.n <= MESSAGE_LIMIT)

如何在SQLAlchemy / Postgres中为每个“group_by”限制N个结果？

1 个答案: