题解 | #平均活跃天数和月活人数#

select
    date_format(submit_time,'%Y%m') as month,
    round(count(distinct uid,date_format(submit_time,'%Y%m%d'))/count(distinct uid),2) as avg_active_days,
    count(distinct uid) as mau
from exam_record
where submit_time is not null and year(submit_time) = 2021
group by date_format(submit_time,'%Y%m');