SELECT round(avg(score),2) AS avg_score
from(    
    SELECT DISTINCT a.rec_user,b.score 
    FROM recommend_tb AS a
    JOIN user_action_tb AS b
    ON a.rec_user = b.user_id AND a.rec_info_l = b.hobby_l
) as C

此题ez,我用了两层SELECT为了去重，因为有个测试用例里面有两次推荐成功题目说的是多次算一次就好