【HIVE高级笔试必备题型】（组内topN、相邻行的值比较问题）求语文大于数学_/_求文科大于理科成绩的学生

Hive SQL练习之成绩分析

数据：[id， 学号，班级，科目，成绩]
1,1,1,yuwen,80
2,1,1,shuxue,85
3,2,1,yuwen,75
4,2,1,shuxue,70
5,3,1,yuwen,86
6,3,1,shuxue,72
7,4,2,yuwen,88
8,4,2,shuxue,99
9,5,2,yuwen,86
10,5,2,shuxue,94
11,6,2,yuwen,56
12,6,2,shuxue,96

题目：

（1）求每个班级前三名的同学（组内topN问题）；

分析：按班级分组，同时取每个同学的平均成绩（或总分）还需要对学生分组。

1）按班级+学生分组，取每个同学平均成绩

select cid, sid, avg(score) avg_score from t_course group by cid, sid order by cid, avg_score desc;
	+------+------+------------+
	| cid  | sid  | avg_score  |
	+------+------+------------+
	| 1    | 1    | 82.5       |
	| 1    | 3    | 79.0       |
	| 1    | 2    | 72.5       |
	| 2    | 4    | 93.5       |
	| 2    | 5    | 90.0       |
	| 2    | 6    | 76.0       |
	+------+------+------------+

	注意，这里因为数据量太少（每个班就三名同学），直接按上面排序就取出了前三名学生。
	这个题实际考察的是多分组（班级+学生），组内TopN（成绩前三名），应该使用dense_rank函数。
		- 请自行回顾row_number、rank、dense_rank三个函数的区别。

	例如取前两名，就得用上dense_rank
	select * from (select cid, sid, avg(score) as avg_score, dense_rank() over (partition by cid order by avg(score) desc) as rank from t_course group by cid, sid  order by cid) t where rank<=2;
		+--------+--------+--------------+---------+
		| t.cid  | t.sid  | t.avg_score  | t.rank  |
		+--------+--------+--------------+---------+
		| 1      | 1      | 82.5         | 1       |
		| 1      | 3      | 79.0         | 2       |
		| 2      | 4      | 93.5         | 1       |
		| 2      | 5      | 90.0         | 2       |
		+--------+--------+--------------+---------+

（2）求语文成绩比数学成绩高的同学，要求使用两种方式完成。

提示：case when，collect_set，concat，concat_set

本人自创方法（也是最简单的方法)

0: jdbc:hive2://bigboss3:10000> select * from(select id,sid,cid,score,lead(score) over() as shuxue from t_course)t where id%2=1 and score>shuxue;
+-------+--------+--------+----------+-----------+
| t.id  | t.sid  | t.cid  | t.score  | t.shuxue  |
+-------+--------+--------+----------+-----------+
| 3     | 2      | 1      | 75       | 70        |
| 5     | 3      | 1      | 86       | 72        |
+-------+--------+--------+----------+-----------+

解析：每个学生都是相邻行科目与成绩不同，使用lead(score) over()把成绩那一列整体向上提升一行，那么奇数行就有该同学语文和数学成绩，这样比起来不就很简单了吗，这个方法最简单了，也是我自己想的，得出答案也是正确的。

方法二：使用collect_set，concat，concat_set，str_to_map

collect_set函数将多行转换为一行，使用concat，concat_set，可以将字段拼接为map格式

0: jdbc:hive2://bigboss3:10000> select * from (select sid,cid,str_to_map(concat_ws(",",collect_set(sc))) as scmap from (select sid,cid,concat(course,":",score) as sc from t_course) t group by sid,cid)s where s.scmap["yuwen"]>s.scmap["shuxue"];
+--------+--------+-------------------------------+
| s.sid  | s.cid  |            s.scmap            |
+--------+--------+-------------------------------+
| 3      | 1      | {"yuwen":"86","shuxue":"72"}  |
| 2      | 1      | {"yuwen":"75","shuxue":"70"}  |
+--------+--------+-------------------------------+
2 rows selected (1.369 seconds)

解析：

1）使用concat方法，将学科和成绩进行拼接
		select sid, concat(course, ":", score) from t_course;
		+------+------------+
		| sid  |    _c1     |
		+------+------------+
		| 1    | yuwen:80   |
		| 1    | shuxue:85  |
		| 2    | yuwen:75   |
		| 2    | shuxue:70  |

		2）使用collect_set函数，每个同学的多行成绩合并到一行
			注意：多行变一行是聚集操作，需要分组
		select sid, collect_set(concat(course, ":", score)) from t_course group by sid;
		+------+---------------------------+
		| sid  |            _c1            |
		+------+---------------------------+
		| 4    | ["yuwen:88","shuxue:99"]  |
		| 6    | ["yuwen:56","shuxue:96"]  |
		| 2    | ["yuwen:75","shuxue:70"]  |
		| 1    | ["yuwen:80","shuxue:85"]  |
		| 3    | ["yuwen:86","shuxue:72"]  |
		| 5    | ["yuwen:86","shuxue:94"]  |
		+------+---------------------------+

		3）转换为map格式，方便获取成绩
			注意：上面获取的["yuwen:88","shuxue:99"]是数组，无法直接转换为map，需要先使用concat_ws将数组转换为字符串
			select sid, concat_ws(",", collect_set(concat(course, ":", score))) from t_course group by sid;
			+------+---------------------+
			| sid  |         _c1         |
			+------+---------------------+
			| 4    | yuwen:88,shuxue:99  |
			| 6    | yuwen:56,shuxue:96  |
			| 2    | yuwen:75,shuxue:70  |

			再将字符串转换为map
			select sid, str_to_map(concat_ws(",", collect_set(concat(course, ":", score)))) as score_map from t_course group by sid;
			+------+-------------------------------+
			| sid  |           score_map           |
			+------+-------------------------------+
			| 4    | {"yuwen":"88","shuxue":"99"}  |
			| 6    | {"yuwen":"56","shuxue":"96"}  |
			| 2    | {"yuwen":"75","shuxue":"70"}  |
			| 1    | {"yuwen":"80","shuxue":"85"}  |
			| 3    | {"yuwen":"86","shuxue":"72"}  |
			| 5    | {"yuwen":"86","shuxue":"94"}  |
			+------+-------------------------------+

		4）通过map访问学科和成绩，并进行过滤
		 select sid, score_map["yuwen"] as yuwen, score_map["shuxue"] as shuxue from (select sid, str_to_map(concat_ws(",", collect_set(concat(course, ":", score)))) as score_map from t_course group by sid) t where score_map["yuwen"]>score_map["shuxue"];
		+------+--------+---------+
		| sid  | yuwen  | shuxue  |
		+------+--------+---------+
		| 2    | 75     | 70      |
		| 3    | 86     | 72      |
		+------+--------+---------+

方法三：case when

这个方法比较巧妙，一般想不出来

select sid, max(yuwen) as score_yw, max(shuxue) as score_sx from (select *, case course when "yuwen" then score else 0 end as yuwen, case course when "shuxue" then score else 0 end as shuxue from t_course) t group by sid  having score_yw>score_sx;
		+------+-----------+-----------+
		| sid  | score_yw  | score_sx  |
		+------+-----------+-----------+
		| 2    | 75        | 70        |
		| 3    | 86        | 72        |
		+------+-----------+-----------+

解析

思路：由于各科成绩在不同行，无法直接进行比较，可以使用case  when新增yuwen、shuxue两列，然后再将各科成绩转换为一行进行比较。
		1）新增yuwen、shuxue两列
		 select *, case course when "yuwen" then score else 0 end as yuwen, case course when "shuxue" then score else 0 end as shuxue from t_course;
			+--------------+---------------+---------------+------------------+-----------------+--------+---------+
			| t_course.id  | t_course.sid  | t_course.cid  | t_course.course  | t_course.score  | yuwen  | shuxue  |
			+--------------+---------------+---------------+------------------+-----------------+--------+---------+
			| 1            | 1             | 1             | yuwen            | 80              | 80     | 0       |
			| 2            | 1             | 1             | shuxue           | 85              | 0      | 85      |
			| 3            | 2             | 1             | yuwen            | 75              | 75     | 0       |
			| 4            | 2             | 1             | shuxue           | 70              | 0      | 70      |
			| 5            | 3             | 1             | yuwen            | 86              | 86     | 0       |
			| 6            | 3             | 1             | shuxue           | 72              | 0      | 72      |
			| 7            | 4             | 2             | yuwen            | 88              | 88     | 0       |
			| 8            | 4             | 2             | shuxue           | 99              | 0      | 99      |
			| 9            | 5             | 2             | yuwen            | 86              | 86     | 0       |
			| 10           | 5             | 2             | shuxue           | 94              | 0      | 94      |
			| 11           | 6             | 2             | yuwen            | 56              | 56     | 0       |
			| 12           | 6             | 2             | shuxue           | 96              | 0      | 96      |
			+--------------+---------------+---------------+------------------+-----------------+--------+---------+

		case course when "yuwen" then score else 0 end as yuwen解析：
			对每一行，当course取值为"yuwen"时，则新增的yuwen字段值为score（语文成绩），否则（course值为shuxue），取值为0；

		2）对每一个学生做分组，目的是将每个学生的成绩显示到一行
		select sid, max(yuwen) as score_yw, max(shuxue) as score_sx from (select *, case course when "yuwen" then score else 0 end as yuwen, case course when "shuxue" then score else 0 end as shuxue from t_course) t group by sid order by sid;
		+------+-----------+-----------+
		| sid  | score_yw  | score_sx  |
		+------+-----------+-----------+
		| 1    | 80        | 85        |
		| 2    | 75        | 70        |
		| 3    | 86        | 72        |
		| 4    | 88        | 99        |
		| 5    | 86        | 94        |
		| 6    | 56        | 96        |
		+------+-----------+-----------+

		3）查找语文成绩更高的学生，注意分组条件过滤使用having，而不是where
		select sid, max(yuwen) as score_yw, max(shuxue) as score_sx from (select *, case course when "yuwen" then score else 0 end as yuwen, case course when "shuxue" then score else 0 end as shuxue from t_course) t group by sid  having score_yw>score_sx;
		+------+-----------+-----------+
		| sid  | score_yw  | score_sx  |
		+------+-----------+-----------+
		| 2    | 75        | 70        |
		| 3    | 86        | 72        |
		+------+-----------+-----------+