题解 | #每天的日活数及新用户占比#

select
    dt,
    count(distinct uid) as dau,
    round(sum(if(rk=1,1,0))/count(distinct uid),2)
from(
    select 
        uid,
        dt,
        rank() over (partition by uid order by dt asc) as rk
    from(
        select
            uid,
            date(in_time) as dt
        from 
            tb_user_log
        union
        select
            uid,
            date(out_time) as dt
        from 
            tb_user_log ) as t1) as t2
group by 1
order by dt asc