Question

我有以下内容：

with t as (
      SELECT advertisable, EXTRACT(YEAR from day) as yy, EXTRACT(MONTH from day) as mon, 
             ROUND(SUM(cost)/1e6) as val
      FROM adcube dac
      WHERE advertisable IN (SELECT advertisable
                                 FROM adcube dac 
                                 GROUP BY advertisable
                                 HAVING SUM(cost)/1e6 > 100
                                )
      GROUP BY advertisable, EXTRACT(YEAR from day), EXTRACT(MONTH from day)
     )
select advertisable, min(yy * 10000 + mon) as yyyymm
from (select t.*,
             (row_number() over (partition by advertisable order by yy, mon) -
              row_number() over (partition by advertisable, val order by yy, mon)
             ) as grp
      from t
     )as foo
group by advertisable, grp, val
having count(*) >= 6 and val = 0
;

这会跟踪停止消费4个月的帐户的激活日期。但是，我想跟踪重新激活日期。因此，如果帐户在4个月后再次开始花费，我可以看到该帐户的新开始日期？

Answer 1

您希望找到val > 0的帐户，并且有4个（或6个）前面的记录为0。

这是一个想法：

计算查询中类似值的组。
为每个组分配一个序号（val_seqnum）。
然后拉出每条记录的先前值和序号。

现在，您需要以下内容为真的记录：

val > 0
prev_val = 0
之前的val_seqnum >= 4（或任何您的门槛）。

以下查询应该这样做（假设t的定义相同）：

select t.*
from (select t.* ,
             lag(val) over (partition by advertisable order by yy, mon) prev_val,
             lag(val_seqnum) over (partition by advertisable order by yy, mon) as prev_val_seqnum
      from (select t.*,
                   row_number() over (partition by advertisable, val, grp order by yy, mon) as val_seqnum
                  ) as grp
            from (select t.*,
                         (row_number() over (partition by advertisable order by yy, mon) -
                          row_number() over (partition by advertisable, val order by yy, mon)
                         ) as grp
                  from t
                 ) t
           ) t
     ) t
where val > 0 and prev_val = 0 and prev_val_seqnum >= 4;

Answer 2

我认为这可以更简单（并且更快）：

SELECT advertisable, ym AS reactivation_ym
FROM (
   SELECT advertisable
        , date_trunc('month', day) AS ym
        , SUM(cost) < 500000       AS asleep
        , count(SUM(cost) < 500000 OR NULL)
                OVER (PARTITION BY advertisable
                      ORDER BY date_trunc('month', day)
                      ROWS BETWEEN 4 PRECEDING AND 1 PRECEDING) AS ct
   FROM   adcube dac
   JOIN  (
      SELECT advertisable
      FROM   adcube
      GROUP  BY 1
      HAVING SUM(cost) > 1e8   -- really 10000000 ?
      ) x USING (advertisable)
   GROUP BY 1, 2
   ) sub
WHERE  NOT asleep
AND    ct = 4;

基于几个假设来填补缺失的信息我在很大程度上解开了你的计算并简化了代码，使其比你原来的更短更快。

计算每个advertisable过去4个月中有多少人的总数cost低于500000.只有当所有4个（现有）月份低于阈值时，该行才有资格。（如果您没有所有月份的行，则需要决定如何处理缺失的行。您的问题中没有相关信息。）

使用count()作为窗口聚合函数和自定义框架。这是最近的相关答案，详细解释如下：

Querying count on daily basis with date constraints over multiple weeks

如何“嵌套”count()和sum()？
它们并不是真正嵌套的。它是一个聚合函数的窗口函数。详细说明：

Get the distinct sum of a joined table column

重新激活SQL

2 个答案: