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
    year (submit_time) = 2021 and submit_time is not null
group by
    month;

用户平均月活跃天 = （用户月活跃天数）/（用户数) ；

用户月活跃天数对 用户数和日期天 去重