重新激活SQL

Question

我有以下幾點：

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 
; 

這會跟蹤站花了4個月的帳戶的激活日期。不過，我想要追蹤重新激活日期。因此，如果帳戶在4個月後再次開始支出，我可以看到該帳戶的新開始日期？

Answer 1

您想查找val > 0以及有4個（或6個）前面0記錄的帳戶。

這裏有一個想法：

• 計算類似值的組爲您的查詢。
• 爲每個組分配一個序號（val_seqnum）。
• 然後拉出每個記錄的前一個值和序列號。

現在，你想要的記錄，其中符合下列條件：

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

以下查詢應該這樣做（假設t含義相同）：

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

我認爲這是可以根本上簡單（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; 

大廈根據一些假設來填補缺失的信息。
我基本上解開了你的計算，簡化了代碼，使它比你的原始代碼更短更快。

• 計算每個advertisable多少的最後4個月總共有cost低於50萬。只有低於閾值的所有4個（現有）個月，該行資格。 （如果您沒有爲所有月份行，你需要決定如何處理缺失行。信息不可用在你的問題。）

使用count()與定製框架窗口聚合函數。下面是最近相關答案有詳細的解釋：

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

你怎麼能 「鳥巢」 count()和sum()？
它們並不真正嵌套。這是一個聚合函數的窗口函數。詳細信息：

• Get the distinct sum of a joined table column