题解 | #平均活跃天数和月活人数#

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