前提
最近在学习hive,碰到了级联求和的问题.经过一番思考学习,现在做些学习笔记.
需求
原始数据表
访客 | 月份 | 访问次数 |
---|---|---|
A | 2015-01 | 5 |
A | 2015-01 | 15 |
B | 2015-01 | 5 |
A | 2015-01 | 8 |
B | 2015-01 | 25 |
A | 2015-01 | 5 |
A | 2015-02 | 4 |
A | 2015-02 | 6 |
B | 2015-02 | 10 |
B | 2015-02 | 5 |
根据上面的数据表输出每个用户每个月份的访问次数,并且每个月统计总的访问次数.最后的输出格式如下
需要输出报表
访客 | 月份 | 月访问总计 | 累计访问总计 |
---|---|---|---|
A | 2015-01 | 33 | 33 |
A | 2015-02 | 10 | 43 |
B | 2015-01 | 30 | 30 |
B | 2015-02 | 15 | 45 |
实现步骤
1.创建hive表
create table t_access_times(username string,month string,cnt int)
row format delimited fields terminated by ',';
2.准备数据 access.log
A,2015-01,5
A,2015-01,15
B,2015-01,5
A,2015-01,8
B,2015-01,25
A,2015-01,5
A,2015-02,4
A,2015-02,6
B,2015-02,10
B,2015-02,5
3.加载数据到表中
load data local inpath '/home/hadoop/access.log' into table t_access_times;
4.自join方式
- 先求每个用户每个月的访问总次数
+-----------+----------+---------+--+
| username | month | cnt |
+-----------+----------+---------+--+
| A | 2015-01 | 33 |
| A | 2015-02 | 10 |
| B | 2015-01 | 30 |
| B | 2015-02 | 15 |
+-----------+----------+---------+--+
- 将月总次数表 自己连接自己(自join)
select A.*,B.* FROM
(select username,month,sum(cnt) as cnt from t_access_times group by username,month) A
inner join
(select username,month,sum(cnt) as cntfrom t_access_times group by username,month) B
on
A.username=B.username
where B.month <= A.month;
+-------------+----------+-----------+-------------+----------+--------
| A.username | A.month | A.cnt| B.username | B.month | B.cnt |
+-------------+----------+-----------+-------------+----------+--------
| A | 2015-01 | 33 | A | 2015-01 | 33 |
| A | 2015-02 | 10 | A | 2015-01 | 33 |
| A | 2015-02 | 10 | A | 2015-02 | 10 |
| B | 2015-01 | 30 | B | 2015-01 | 30 |
| B | 2015-02 | 15 | B | 2015-01 | 30 |
| B | 2015-02 | 15 | B | 2015-02 | 15 |
+-------------+----------+-----------+-------------+----------+--------
刚开始这里不是很明白为什么加上where B.month >= A.month
的条件,这样有什么意义?其实这是为后面的统计做准备.
现在来讲讲这个自join是怎么产生这样的数据的.
hive的表连接我没有研究过,这里暂时我用mysql的连接来举例说明.我姑且认为它们的实现原理的是一样的.
- 从表A中读入一行数据R;
- 从数据行R中,取出username字段和where条件到B表中去查找;
- 在B表中找到满足条件的行,跟R组成一行,作为结果集的一部分;
- 重复执行步骤1到3,直到表A的末尾循环结束;
在这里,两个表都做了一次全表扫描,所以总的扫描行数是 4 + 4 = 8;
内存中的判断次数是 4 * 4 = 16;
- 最终的sql语句
select A.username,A.month,max(A.cnt) as cnt,sum(B.cnt) as accumulate
from
(select username,month,sum(cnt) as cntfrom t_access_times group by username,month) A
inner join
(select username,month,sum(cnt) as cntfrom t_access_times group by username,month) B
on
A.username=B.username
where B.month <= A.month
group by A.username,A.month
order by A.username,A.month;
--最终结果为:
+-------------+----------+---------+-------------+--+
| A.username | A.month | cnt| accumulate |
+-------------+----------+---------+-------------+--+
| A | 2015-01 | 33 | 33 |
| A | 2015-02 | 10 | 43 |
| B | 2015-01 | 30 | 30 |
| B | 2015-02 | 15 | 45 |
+-------------+----------+---------+-------------+--+
5.窗口函数
还有一种方式也可以实现需求,那就是窗口函数
select
t.username,
t.month,
t.cnt,
sum(t.cnt) over(partition by t.username order by t.username,
t.month rows between unbounded preceding and current row) as accumlate
from(
select
username,month,
sum(cnt) as cnt
from t_access_times group by username,month) t
;
--最终结果为:
+-------------+----------+---------+-------------+--+
| A.username | A.month | cnt| accumulate |
+-------------+----------+---------+-------------+--+
| A | 2015-01 | 33 | 33 |
| A | 2015-02 | 10 | 43 |
| B | 2015-01 | 30 | 30 |
| B | 2015-02 | 15 | 45 |
+-------------+----------+---------+-------------+--+
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