如何使用 sqlalchemy 计算平均数？答案

【问题标题】：How to take avg on count using sqlalchemy?如何使用 sqlalchemy 计算平均数？
【发布时间】：2020-05-27 03:33:46
【问题描述】：

如何在 SQLAlchemy 中获取平均计数？我尝试了很多东西，但都失败了。代码：

month = sqlalchemy.func.date_trunc('month', Complaint.date)
records_count = sqlalchemy.sql.func.count(ClientGroupRecord.id)
complained_clients = session.query(Friend, records_count.label('count'), month). \
    filter(Friend.friend_id == friend_id). \
    join(Complaint, Complaint.friend == Friend.friend_id). \
    join(ClientGroup, Complaint.client_group == ClientGroup.client_group_id). \
    join(ClientGroupRecord, ClientGroup.client_group_id == ClientGroupRecord.client_group_id). \
    join(Client, Client.client_id == ClientGroupRecord.client_id). \
    group_by(month, Friend.friend_id)

records_avg = sqlalchemy.sql.func.avg(records_count)

result = ???

在result 中，我不需要Friend，但希望在列上使用avg，该列标记为count，按month 列分组。

【问题讨论】：

如果您提供一些您所做的尝试，通常有助于获得正确的答案。
相关：stackoverflow.com/questions/33826441/…

标签： python sql database orm sqlalchemy

【解决方案1】：

一个选项是使用窗口函数，如果您的 DBMS 支持它们 - 基于 date_trunc 的使用我假设您使用的是 PostgreSQL。这样做的原因是在 GROUP BY 生成组行之后评估窗口函数。所以将计数包装在AVG(...) OVER () 中，其中窗口是整个结果集：

from sqlalchemy import func

month = func.date_trunc('month', Complaint.date)
records_count = func.count(ClientGroupRecord.id)
# Create a window of the whole set of results
records_avg = func.avg(records_count).over()

complained_clients = session.query(records_count.label('count'),
                                   records_avg.label('avg'),
                                   month.label('month')). \
    select_from(Friend). \
    join(Complaint, Complaint.friend == Friend.friend_id). \
    join(ClientGroup, Complaint.client_group == ClientGroup.client_group_id). \
    join(ClientGroupRecord, ClientGroup.client_group_id == ClientGroupRecord.client_group_id). \
    join(Client, Client.client_id == ClientGroupRecord.client_id). \
    filter(Friend.friend_id == friend_id). \
    group_by(month). \
    all()

请注意，Friend 已从 query(...) 移动到显式用作与 Query.select_from() 的第一个连接的左侧。即使您只选择具有特定 ID 的朋友，也无需在 GROUP BY 子句中使用 Friend.id。

如果您对每月计数根本不感兴趣，而只想要平均值，那么更传统的subquery approach 也一样好：

from sqlalchemy import func

month = func.date_trunc('month', Complaint.date)
records_count = func.count(ClientGroupRecord.id)

complained_clients = session.query(records_count.label('count')). \
    select_from(Friend). \
    join(Complaint, Complaint.friend == Friend.friend_id). \
    join(ClientGroup, Complaint.client_group == ClientGroup.client_group_id). \
    join(ClientGroupRecord, ClientGroup.client_group_id == ClientGroupRecord.client_group_id). \
    join(Client, Client.client_id == ClientGroupRecord.client_id). \
    filter(Friend.friend_id == friend_id). \
    group_by(month). \
    subquery()

result = session.query(func.avg(complained_clients.c.count)).scalar()

...或使用窗口方法：

from sqlalchemy import func

month = func.date_trunc('month', Complaint.date)
records_count = func.count(ClientGroupRecord.id)
# Create a window of the whole set of results
records_avg = func.avg(records_count).over()

# Windows are evaluated for each row, but here we have a single window spanning
# the entire result, so the use of DISTINCT collapses this to a single value.
# Knowing what the query does LIMIT 1 / FETCH FIRST 1 ROW ONLY would work as well.
result = session.query(records_avg.label('avg')). \
    select_from(Friend). \
    join(Complaint, Complaint.friend == Friend.friend_id). \
    join(ClientGroup, Complaint.client_group == ClientGroup.client_group_id). \
    join(ClientGroupRecord, ClientGroup.client_group_id == ClientGroupRecord.client_group_id). \
    join(Client, Client.client_id == ClientGroupRecord.client_id). \
    filter(Friend.friend_id == friend_id). \
    group_by(month). \
    distinct(). \
    scalar()

【讨论】：