sql - 使用 SQL 查询识别趋势

Question

我有一个包含一组对象 ID、数值和日期的表（我们称之为数据）。我想确定在过去 X 分钟（例如一小时）内其值具有正趋势的对象。

示例数据：

entity_id | value | date

1234      | 15    | 2014-01-02 11:30:00

5689      | 21    | 2014-01-02 11:31:00

1234      | 16    | 2014-01-02 11:31:00

我尝试查看类似的问题，但不幸的是没有找到任何帮助...

Answer 1

您启发了我在 SQL Server 中实现线性回归。这可以为 MySQL/Oracle/Whatever 修改，而不会有太多麻烦。这是确定每个 entity_id 一小时内趋势的最佳数学方法，它将仅选择具有积极趋势的那些。

它实现了此处列出的计算 B1hat 的公式：https ://en.wikipedia.org/wiki/Regression_analysis#Linear_regression

create table #temp
(
    entity_id int,
    value int,
    [date] datetime
)

insert into #temp (entity_id, value, [date])
values
(1,10,'20140102 07:00:00 AM'),
(1,20,'20140102 07:15:00 AM'),
(1,30,'20140102 07:30:00 AM'),
(2,50,'20140102 07:00:00 AM'),
(2,20,'20140102 07:47:00 AM'),
(3,40,'20140102 07:00:00 AM'),
(3,40,'20140102 07:52:00 AM')

select entity_id, 1.0*sum((x-xbar)*(y-ybar))/sum((x-xbar)*(x-xbar)) as Beta
from
(
    select entity_id,
        avg(value) over(partition by entity_id) as ybar,
        value as y,
        avg(datediff(second,'20140102 07:00:00 AM',[date])) over(partition by entity_id) as xbar,
        datediff(second,'20140102 07:00:00 AM',[date]) as x
    from #temp
    where [date]>='20140102 07:00:00 AM' and [date]<'20140102 08:00:00 AM'
) as Calcs
group by entity_id
having 1.0*sum((x-xbar)*(y-ybar))/sum((x-xbar)*(x-xbar))>0

Answer 2

如果有人在 Mysql 中需要这个，这是适合我的代码。

datapoint | plays | status_time 

1234      | 15    | 2014-01-02 11:30:00

5689      | 21    | 2014-01-02 11:31:00

1234      | 16    | 2014-01-02 11:31:00

select datapoint, 1.0*sum((x-xbar)*(y-ybar))/sum((x-xbar)*(x-xbar)) as Beta
from
(
     select datapoint,
        avg(plays) over(partition by datapoint) as ybar,
        plays as y,
        avg(TIME_TO_SEC(TIMEDIFF('2021-03-22 21:00:00', status_time))) over(partition by datapoint) as xbar,
        TIME_TO_SEC(TIMEDIFF('2021-03-22 21:00:00', status_time)) as x
    from aggregate_datapoints
    where status_time BETWEEN'2021-03-22 21:00:00' and '2021-03-22 22:00:00'
and type = 'topContent') as calcs
group by datapoint
having 1.0*sum((x-xbar)*(y-ybar))/sum((x-xbar)*(x-xbar))>0