SQL面试题-留存率

留存率是衡量用户质量的最重要指标之一，因此计算用户留存率是用户数据分析中必须掌握的技能之一，同样也成为了面试经典sql之一。
留存率指标中，通常需要关注次日留存、3日留存、7日留存和月留存。对新增用户而言，需要关注更细颗粒度的数据，也就是7日内每天的留存率。

代码实现

select
    dt
    ,count( if(id=lead_id  and datediff(lead_dt,dt) =1 ,id, null ) ) as `1日留存`
    ,count( if(id=lead_id7 and datediff(lead_dt7,dt)=7 ,id, null ) ) as `7日留存`
from
(
    select
        id 
        ,dt
        ,lead(dt,1) over(partition by id order by dt asc ) as lead_dt
        ,lead(id,1) over(partition by id order by dt asc ) as lead_id
        ,lead(dt,7) over(partition by id order by dt asc ) as lead_dt7
        ,lead(id,7) over(partition by id order by dt asc ) as lead_id7
    from  
    (
        select 'slm' as id, '2018-12-26' as dt
        union all select 'slm' as id, '2018-12-27' as dt
        union all select 'slm' as id, '2018-12-28' as dt
        union all select 'hh ' as id, '2018-12-26' as dt
        union all select 'hh ' as id, '2018-12-28' as dt 
    ) a
) b
group by dt 
order by dt