1

我有一张这样的桌子:

id |    job     | school  |
1  | programmer | school1 |
2  | programmer | school1 |
3  | programmer | school2 |
4  |     pm     | school3 |
5  |     pm     | school2 |
6  |     pm     | school3 |

我想做以下事情:

  1. 按工作分组
  2. 获取学校列表和计数,像这样[(school1, 2), (school2, 1)]
  3. 学校列表按计数排序,所以不能是 [(school1, 1), (school1, 2)]

该示例的结果是:

programmer  |  [(school1, 2), (school2, 1)]
    pm      |  [(school3, 2), (school2, 1)]
4

2 回答 2

1

我们不能在 hive 的 Collection (collect_set) 中拥有 Map(因为在 collect_set 中只允许使用原始数据类型)。

这 2 个查询将提供您要查找的内容(两者都相同,除了一个涉及地图,其他不涉及)

CREATE EXTERNAL TABLE job_test(
  id string,
  job string,
  school string )
ROW FORMAT DELIMITED
FIELDS TERMINATED BY ','
LINES TERMINATED BY '\n'
STORED AS TEXTFILE
LOCATION  '/user/test/job.txt';

SELECT b.job, collect_set(concat_ws(':',map_keys(b.school_map),map_values(b.school_map))) as school_cnt
FROM
(
    SELECT a.job, map(a.school,a.cnt) as school_map
    FROM
    ( 
         SELECT job,
                school, 
                cast(count(1) as string) as cnt
         FROM job_test
         GROUP BY 
                job,
                school
    )a
)b
GROUP BY b.job;

SELECT a.job, collect_set(concat_ws(':',a.school,a.cnt)) as school_cnt
FROM
(
     SELECT job,
            school,
            cast(count(1) as string) as cnt
     FROM job_test
     GROUP BY 
            job,
            school
)a
GROUP BY a.job;

希望这可以帮助 :)

于 2015-05-19T13:53:21.127 回答
1

只需添加Brickhouse jar 并创建一个collect()函数

add jar ./brickhouse-0.7.1.jar;
create temporary function collect as 'brickhouse.udf.collect.CollectUDAF';

select job
  , collect(school, c) school_count_map
from (
  select *
  from (
    select job, school
      , count( * ) c
    from table
    group by job, school ) x
  order by job, c desc) y
group by job
于 2015-05-21T04:33:16.730 回答