select device_id,ym ,sum(question_cnt) over(partition by device_id order by ym) as sum_cnt -- 截至当月的每人练题总数 ,round(sum(question_cnt) over(partition by device_id order by ym rows 2 preceding)/count(ym) over(partition by device_id order by ym rows 2 preceding),2) as avg3_cnt ,sum(question_cnt)over(order...