题解 | #平均活跃天数和月活人数#

select
    concat (year (submit_time), "0", month (submit_time)) month,
    round(
        count(distinct uid, left (submit_time, 10)) / count(distinct uid),
        2
    ) avg_active_days,
    count(distinct uid) mau
from
    exam_record
where
    year (submit_time) = 2021
group by
    concat (year (submit_time), "0", month (submit_time))