select round(sum(score)/count(distinct user_id),3) as avg_score
from user_action_tb 
where user_id in
(select user_id 
from user_action_tb as tb1
join recommend_tb  as tb2 on 
tb1.user_id = tb2.rec_user and tb1.hobby_l = tb2.rec_info_l
)

用连接和子查询