重新激活SQL

Question

我有以下几点：

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 
; 

这会跟踪站花了4个月的帐户的激活日期。不过，我想要追踪重新激活日期。因此，如果帐户在4个月后再次开始支出，我可以看到该帐户的新开始日期？

Answer 1

您想查找val > 0以及有4个（或6个）前面0记录的帐户。

这里有一个想法：

• 计算类似值的组为您的查询。
• 为每个组分配一个序号（val_seqnum）。
• 然后拉出每个记录的前一个值和序列号。

现在，你想要的记录，其中符合下列条件：

• val > 0
• prev_val = 0
• 以前val_seqnum >= 4（或任何你的阈值）。

以下查询应该这样做（假设t含义相同）：

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

我认为这是可以根本上简单（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; 

大厦根据一些假设来填补缺失的信息。
我基本上解开了你的计算，简化了代码，使它比你的原始代码更短更快。

• 计算每个advertisable多少的最后4个月总共有cost低于50万。只有低于阈值的所有4个（现有）个月，该行资格。 （如果您没有为所有月份行，你需要决定如何处理缺失行。信息不可用在你的问题。）

使用count()与定制框架窗口聚合函数。下面是最近相关答案有详细的解释：

• Querying count on daily basis with date constraints over multiple weeks

你怎么能 “鸟巢” count()和sum()？
它们并不真正嵌套。这是一个聚合函数的窗口函数。详细信息：

• Get the distinct sum of a joined table column