Select JSON 数组中元素的前 3 名

Question

源数据为

user_id video_interest
1 [{"category":"a","score":1},{"category":"b","score":2},{"category":"c","score":3},{"category":"d","score":4}]
2 [{"category":"e","score":1},{"category":"f","score":2},{"category":"g","score":-3}]

输出为

user_id video_interest_top3
1 [{"category":"d","score":4},{"category":"c","score":3},{"category":"b","score":2}]
2 [{"category":"f","score":2},{"category":"e","score":1}]

我需要过滤score>0，然后select每个user_id的top3video_interest按照score

降序排列

Answer 1

分解JSON数组，提取分数，计算每个用户的最大分数（如果需要按分数desc排序最终输出）和row_number按分数过滤前3名，再次收集数组并连接如有必要，将其添加到 STRING。请参阅代码中的注释。我添加了排序数组和整个输出，因为最初不清楚到底应该排序什么：数组或最终输出，如果不需要，请删除 max_score 排序。

演示：

with mytable as (
select stack(2,  
1,'[{"category":"a","score":1},{"category":"b","score":2},{"category":"c","score":3},{"category":"d","score":4}]',
2,'[{"category":"e","score":1},{"category":"f","score":2},{"category":"g","score":-3}]'
) as (user_id,video_interest)
)

select --collect array and convert to JSON string
      user_id, max_score, concat('[',concat_ws(',',collect_list(category_score)),']') as video_interest
from
(
select user_id, category_score, max_score, score
from
(  
select --extract score, filter and sort
      user_id, vi.category_score, get_json_object(vi.category_score, '$.score') as score,
      row_number() over(partition by user_id order by get_json_object(vi.category_score, '$.score') desc) rn, 
      max(get_json_object(vi.category_score, '$.score')) over (partition by user_id) max_score
from
(--prepare for exploding array
select user_id, regexp_replace(regexp_replace(video_interest,'\[|\]',''), --remove []
                          '\},\{', '},,,{') as video_interest --replace , between array elements with ,,, to split
  from mytable
)s 
--split and explode
lateral view outer explode(split(video_interest,',,,')) vi as category_score
where get_json_object(vi.category_score, '$.score')>0
)s
where rn<=3 --filter top 3
distribute by user_id sort by score desc --Sort collection, remove if not necessary
)s
group by user_id, max_score
order by max_score desc --Sorting users by max_score desc, remove if not necessary

结果：

user_id max_score   video_interest
1       4           [{"category":"d","score":4},{"category":"c","score":3},{"category":"b","score":2}]
2       2           [{"category":"f","score":2},{"category":"e","score":1}]

Answer 2

首先，我展开 video_interest 并将其类别和评分设为单个字段。其次，我使用 row_number() 函数按分数（降序）按 user_id 顺序进行分区，然后使每一行都标有它们在组中的顺序并过滤 order<=3 最后，我使用 collect_list() 将它们简单地收集为一个列表，因为它们在使用 row_number

时是有序的

select  user_id,
            collect_list(pos) as first_video_interest_top3
from    (
            select  user_id,
                    category,
                    score,
                    pos,
                    row_number() over(
                    partition by
                                user_id
                    order by
                                score desc
                    ) rNum
            from    (
                        select  user_id,
                                pos.category,
                                pos.score,
                                pos
                        from    myData
                        lateral view explode(video_interest) e as pos
                        ) t1
            where   score > 0
            ) t2
where   rNum <= 3
group by
            user_id

Select JSON 数组中元素的前 3 名

Select top 3 by element in JSON arrays

sql

arrays

json

hive

hiveql