sql用户保留计算

我在雅典娜有这样一个表

记录

，一个用户一个月一行：

month,   id
2020-05   1
2020-05   2
2020-05   5
2020-06   1
2020-06   5
2020-06   6

需要计算

百分比=（上个月和本月的用户数）/（上个月的总用户数）。

像上面的例子一样，5月和6月1日都有用户，5月总共有3个用户，这应该计算出2/3*100的百分比

with monthly_mau AS 
    (SELECT  month as mauMonth,
         date_format(date_add('month',1,cast(concat(month,'-01') AS date)), '%Y-%m') AS nextMonth,
         count(distinct userid) AS monthly_mau
    FROM records
    GROUP BY  month
    ORDER BY  month),
 
 retention_mau AS 
    (SELECT 
         month,
         count(distinct useridLeft) AS retention_mau
    FROM ( 
        (SELECT 
         userid as useridLeft,month as monthLeft,
         date_format(date_add('month',1,cast(concat(month,'-01') AS date)), '%Y-%m') AS nextMonth
        FROM records ) AS prior
        INNER JOIN 
            (SELECT 
         month ,
         userid 
            FROM records ) AS current
                ON 
                    prior.useridLeft = current.userid
                    AND prior.nextMonth = current.month )
     WHERE userid is not null
     GROUP BY   month
     ORDER BY   month )
 
SELECT *, cast(retention_mau AS double)/cast(monthly_mau AS double)*100 AS retention_mau_percentage 
FROM monthly_mau as m
    INNER JOIN monthly_retention_mau AS r
        ON m.nextMonth = r.month 
 order by  r.month     

---

这给了我100%的百分比，这是不对的。有什么想法吗？

嗯。假设每个用户每月有一行，则可以使用窗口函数和条件聚合：

select month, count(*) as num_users,
       sum(case when prev_month = dateadd('month', -1, month) then 1 else 0 end) as both_months
from (select r.*,
             cast(concat(month, '-01') AS date) as month_date,
             lag(cast(concat(month, '-01') AS date)) over (partition by id order by month) as prev_month_date
      from records r
     ) r
group by month;