题解 | #平均活跃天数和月活人数#

select
    date_format (start_time, '%Y%m') as month,
    round(count(distinct uid, date_format (start_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 (start_time, '%Y%m')