玩了至少 3 节后 3 天内又回来的用户数量？答案

【问题标题】：Number of user that came back within 3 days after playing at least three sessions?玩了至少 3 节后 3 天内又回来的用户数量？
【发布时间】：2018-03-06 17:21:34
【问题描述】：

我有包含用户、事件日期和会话的数据。我想区分至少有 3 个会话并在 3 天内返回新会话的用户。

    user  eventdate   session
    A      2018-02-05   1
    A      2018-02-05   2
    A      2018-02-06   3 
    A      2018-02-10   4

输出完成了 3 次会话然后在 3 天内返回第四次会话的用户。

我尝试了以下查询，但它没有给我所需的答案。

 SELECT distinct user, MIN(eventdate) startdate, MAX(eventdate) enddate
FROM (SELECT user, eventdate
      FROM (SELECT user, eventdate
              FROM tablename
             where datediff(startdate,enddate)<=3
             ORDER BY user, eventdate) where sessions>=3) t
 GROUP BY user
 ORDER BY user, startdate;

我知道查询有很多问题，但我根本无法弄清楚如何继续前进。有什么建议吗？

【问题讨论】：

预期输出是什么？在上述情况下，A 的所有数据都会出现。你的查询编译了吗？

标签： sql google-bigquery legacy-sql

【解决方案1】：

以下是 BigQuery 标准 SQL

#standardSQL
SELECT *
FROM (
  SELECT 
    user, eventdate, sessions_in_a_day, 
    SUM(sessions_in_a_day) OVER(PARTITION BY user ORDER BY eventdate ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING) total_sessions_before, 
    DATE_DIFF(eventdate, LAG(eventdate) OVER(PARTITION BY user ORDER BY eventdate), DAY) delay
  FROM (
    SELECT user, eventdate, COUNT(1) sessions_in_a_day 
    FROM t
    GROUP BY user, eventdate
  )
)
WHERE total_sessions_before >= 3
AND delay <= 3
-- ORDER BY user, eventdate

您可以使用虚拟数据测试/玩上面的内容

#standardSQL
WITH t AS (
  SELECT 'A' user, DATE '2018-02-05' eventdate, 1 session UNION ALL
  SELECT 'A', DATE '2018-02-05', 2 UNION ALL
  SELECT 'A', DATE '2018-02-06', 3 UNION ALL
  SELECT 'A', DATE '2018-02-06', 4 UNION ALL
  SELECT 'A', DATE '2018-02-09', 5 UNION ALL
  SELECT 'A', DATE '2018-02-09', 6 UNION ALL
  SELECT 'A', DATE '2018-02-10', 7 UNION ALL 
  SELECT 'A', DATE '2018-02-13', 8 
)
SELECT *
FROM (
  SELECT 
    user, eventdate, sessions_in_a_day, 
    SUM(sessions_in_a_day) OVER(PARTITION BY user ORDER BY eventdate ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING) total_sessions_before, 
    DATE_DIFF(eventdate, LAG(eventdate) OVER(PARTITION BY user ORDER BY eventdate), DAY) delay
  FROM (
    SELECT user, eventdate, COUNT(1) sessions_in_a_day 
    FROM t
    GROUP BY user, eventdate
  )
)
WHERE total_sessions_before >= 3
AND delay <= 3
ORDER BY user, eventdate

结果是

Row user    eventdate   sessions_in_a_day   total_sessions_before   delay    

1   A       2018-02-09  2                   4                       3    
2   A       2018-02-10  1                   6                       1    
3   A       2018-02-13  1                   7                       3

使用 WHERE 子句，您可以“调整”到您需要的任何情况
在上面的示例中，您仅显示在接下来的 3 天内到达下一个会话之前至少有 3 个会话的用户如果您只对那些恰好有 3 个会话并达到第四个会话的人感兴趣 - 您可以添加相应的过滤器

【讨论】：

【解决方案2】：

WITH Sess AS
(
select user, session
from tablename
group by  user
HAVING count(session) >= 3
)

select user
from tablename join Sess on tablename.session = Sess.session
group by user
having (datediff(day, min(eventdate), Max(eventdate)) <=3) 
and (min(eventdate) <> Max(eventDate))

【讨论】：