常见大数据面试SQL-分组连续累积计数

数据仓库晨曦

发布于 2024-09-10 21:04:33

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文章被收录于专栏：数据仓库技术数据仓库技术

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一、题目

有temp表包含A，B两列，请使用SQL对该B列进行处理，形成C列，按A列顺序，B列值不变，则C列累计计数，C列值变化，则C列重新开始计数，期望结果如下

样例数据

+-------+----+
|   a   | b  |
+-------+----+
| 2010  | 1  |
| 2011  | 1  |
| 2012  | 1  |
| 2013  | 0  |
| 2014  | 0  |
| 2015  | 1  |
| 2016  | 1  |
| 2017  | 1  |
| 2018  | 0  |
| 2019  | 0  |
+-------+----+

期望结果

+-------+----+----+
|   a   | b  | c  |
+-------+----+----+
| 2010  | 1  | 1  |
| 2011  | 1  | 2  |
| 2012  | 1  | 3  |
| 2013  | 0  | 1  |
| 2014  | 0  | 2  |
| 2015  | 1  | 1  |
| 2016  | 1  | 2  |
| 2017  | 1  | 3  |
| 2018  | 0  | 1  |
| 2019  | 0  | 2  |
+-------+----+----+

二、分析

本题是连续问题的变种，在判断连续的基础上进行累积计数。连续问题解决方案参考一文搞懂连续问题
本题考点相对较多，连续问题本身已经较难，会涉及到lag函数、sum()over(order by) 进行累积求和、连续数据处理的技巧，本题在连续的基础上又考察了count(*)over(order by )的累积计数。

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

三、SQL

1.lag()判断是否连续

使用lag判断是否连续，并对连续打标为0，不连续打标为1

执行SQL

select a,
       b,
       if(b = lag(b,1) over (order by a asc), 0, 1) as is_conn
from t19_temp

SQL结果

+-------+----+----------+
|   a   | b  | is_conn  |
+-------+----+----------+
| 2010  | 1  | 1        |
| 2011  | 1  | 0        |
| 2012  | 1  | 0        |
| 2013  | 0  | 1        |
| 2014  | 0  | 0        |
| 2015  | 1  | 1        |
| 2016  | 1  | 0        |
| 2017  | 1  | 0        |
| 2018  | 0  | 1        |
| 2019  | 0  | 0        |
+-------+----+----------+

2.计算分组id

使sum()over(order by )方式计算出连续的分组id:conn_group_id

执行SQL

select a,
       b, 
       sum(is_conn) over (order by a asc) as conn_group_id
from (select a, 
             b,
             if(b = lag(b, 1) over (order by a asc), 0, 1) as is_conn
      from t19_temp
      ) t

SQL结果

+-------+----+----------------+
|   a   | b  | conn_group_id  |
+-------+----+----------------+
| 2010  | 1  | 1              |
| 2011  | 1  | 1              |
| 2012  | 1  | 1              |
| 2013  | 0  | 2              |
| 2014  | 0  | 2              |
| 2015  | 1  | 3              |
| 2016  | 1  | 3              |
| 2017  | 1  | 3              |
| 2018  | 0  | 4              |
| 2019  | 0  | 4              |
+-------+----+----------------+

**3.按照分组id分组，count(*)over(order by) 计数，得出最后结果**

执行SQL

select a,
       b,
       count(1) over (partition by conn_group_id order by a asc) as c
from (select a,
             b,
             sum(is_conn) over (order by a asc) as conn_group_id
      from (select
                a,
                b,
                if(b = lag(b, 1) over (order by a asc), 0, 1) as is_conn
            from db_interview_cj.t19_temp
            ) t
      ) tt

SQL结果

+-------+----+----+
|   a   | b  | c  |
+-------+----+----+
| 2010  | 1  | 1  |
| 2011  | 1  | 2  |
| 2012  | 1  | 3  |
| 2013  | 0  | 1  |
| 2014  | 0  | 2  |
| 2015  | 1  | 1  |
| 2016  | 1  | 2  |
| 2017  | 1  | 3  |
| 2018  | 0  | 1  |
| 2019  | 0  | 2  |
+-------+----+----+

四、建表语句和数据插入

--建表语句
CREATE TABLE IF NOT EXISTS t19_temp
(
    a    string, -- 用户id
    b    bigint -- 登陆日期
)
    COMMENT 'temp';
--插入数据
INSERT INTO t19_temp VALUES
('2010',1),
('2011',1),
('2012',1),
('2013',0),
('2014',0),
('2015',1),
('2016',1),
('2017',1),
('2018',0),
('2019',0);

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原始发表：2024-09-01，如有侵权请联系 cloudcommunity@tencent.com 删除

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