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'
group by date_format(submit_time,'%Y%m')