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常见题目
12.常见大数据面试SQL-查询前2大和前2小用户并有序拼接

常见大数据面试SQL-查询前2大和前2小用户并有序拼接

一、题目

有用户账户表,包含年份,用户id和值,请按照年份分组,取出值前两小和前两大对应的用户id,**注意:**需要保持值最小和最大的用户id排首位。

样例数据

+-------+----------+--------+
| year  | user_id  | value  |
+-------+----------+--------+
| 2022  | A        | 30     |
| 2022  | B        | 10     |
| 2022  | C        | 20     |
| 2023  | A        | 40     |
| 2023  | B        | 50     |
| 2023  | C        | 20     |
| 2023  | D        | 30     |
+-------+----------+--------+

期望结果

+-------+-----------------+-----------------+
| year  | max2_user_list  | min2_user_list  |
+-------+-----------------+-----------------+
| 2022  | A,C             | B,C             |
| 2023  | B,A             | C,D             |
+-------+-----------------+-----------------+

二、分析

属于取最大最小记录的升级版,最大难点在于拼用户要保证有序。

维度评分
题目难度⭐️⭐️⭐️⭐️
题目清晰度⭐️⭐️⭐️⭐️⭐️
业务常见度⭐️⭐️⭐️⭐️

三、SQL

1.row_number函数根据年份分组,value正排和倒排得到两个序列

使用row_number函数根据年份分组,根据value正排得到 asc_rn用于取出value最小两行记录,根据value倒叙得到desc_rn用于取出最大两行记录

执行SQL

select user_id
     , year
     , value
     , row_number() over (partition by year order by value desc) as desc_rn
     , row_number() over (partition by year order by value)      as rn
from t12_amount

查询结果

+----------+-------+--------+----------+-----+
| user_id  | year  | value  | desc_rn  | rn  |
+----------+-------+--------+----------+-----+
| B        | 2022  | 10     | 3        | 1   |
| C        | 2022  | 20     | 2        | 2   |
| A        | 2022  | 30     | 1        | 3   |
| C        | 2023  | 20     | 4        | 1   |
| D        | 2023  | 30     | 3        | 2   |
| A        | 2023  | 40     | 2        | 3   |
| B        | 2023  | 50     | 1        | 4   |
+----------+-------+--------+----------+-----+

2.根据年份分组,取出value最大user_id,第二大user_id,最小user_id,第二小user_id

根据年份分组,取出每年最大、第二大,最小、第二小用户ID。使用if对desc_rn,rn进行判断,对符合条件的数据取出user_id,其他去null,然后使用聚合函数取出结果。

执行SQL

select year,
       max(if(desc_rn = 1, user_id, null)) as max1_user_id,
       max(if(desc_rn = 2, user_id, null)) as max2_user_id,
       max(if(rn = 1, user_id, null))      as min1_user_id,
       max(if(rn = 2, user_id, null))      as min2_user_id
from (select user_id
           , year
           , value
           , row_number() over (partition by year order by value desc) as desc_rn
           , row_number() over (partition by year order by value)      as rn
      from t12_amount) t1
group by year

查询结果

+-------+---------------+---------------+---------------+---------------+
| year  | max1_user_id  | max2_user_id  | min1_user_id  | min2_user_id  |
+-------+---------------+---------------+---------------+---------------+
| 2022  | A             | C             | B             | C             |
| 2023  | B             | A             | C             | D             |
+-------+---------------+---------------+---------------+---------------+

3.按照顺序拼接,得到最终结果

按照题目要求,进行字符拼接

  • 拼接max1_user_id、max2_user_id为max2_list;
  • 拼接min1_user_id、min2_user_id为min2_list;

执行SQL

select year,
       concat(max(if(desc_rn = 1, user_id, null)), ',',
              max(if(desc_rn = 2, user_id, null))) as max2_user_list,
       concat(max(if(rn = 1, user_id, null)), ',',
              max(if(rn = 2, user_id, null)))      as min2_user_list
from (select user_id
           , year
           , value
           , row_number() over (partition by year order by value desc) as desc_rn
           , row_number() over (partition by year order by value)      as rn
      from t12_amount) t1
group by year

查询结果

+-------+-----------------+-----------------+
| year  | max2_user_list  | min2_user_list  |
+-------+-----------------+-----------------+
| 2022  | A,C             | B,C             |
| 2023  | B,A             | C,D             |
+-------+-----------------+-----------------+

四、建表语句和数据插入

--建表语句
create table if not exists t12_amount
(
    year    string,
    user_id string,
    value   bigint
);
 
--插入数据
 
insert into t12_amount(year, user_id, value)
values ('2022', 'A', 30),
       ('2022', 'B', 10),
       ('2022', 'C', 20),
       ('2023', 'A', 40),
       ('2023', 'B', 50),
       ('2023', 'C', 20),
       ('2023', 'D', 30)

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