第一次速度击败了100%的人，只用总和处于人数就可以，没用到题解的在对id进行商品的聚合，所以会快很多 用了 （） in （select）来匹配每个用户的第一次订单，

select round(amount/num,1) ,
       round((cost-amount)/num,1)
from
(select sum(price*cnt) as cost,
       count(distinct uid) as num,
       sum(distinct total_amount) as amount
from  tb_order_detail de 
left join  tb_product_info info  using(product_id)
left join tb_order_overall ov  using(order_id)
where (uid ,event_time) in (select uid,min(event_time)
                    from tb_order_overall  group by uid)
       and left(event_time,7)="2021-10" and status=1) tmp