题解 | #平均活跃天数和月活人数#

select
    replace(substr(start_time, 1, 7), '-', '')       as month,
    round(count(distinct uid, if(score is not null, substr(start_time, 1, 10), null)) /
    count(distinct if(score is not null, uid, null)),2) as avg_active_days,
    count(distinct if(score is not null, uid, null)) as mau
from
    exam_record
where
    year(start_time) = '2021'
group by
    replace(substr(start_time, 1, 7), '-', '');