sql - 如何在 hive 中通过 collect_set() 操作使用 order by

Question

在表 1 中，我有 customer_id、item_id 和 item_rank(根据一些销售额的项目排名)。我想为每个 customer_id 收集一个项目列表，并根据 item_rank 排列它们。

Customer_id  item_id rank_item
  23            2      3
  23            2      3
  23            4      2
  25            5      1
  25            4      2

我期望的输出是

Customer_id    item_list
  23             4,2
  25             5,4

我使用的代码是

 SELECT
    customer_id,
    concat_ws(',',collect_list (string(item_id))) AS item_list
FROM
    table1
GROUP BY
    customer_id
ORDER BY
    item_rank

Answer 1

可以使用子查询得到(customer_id, item_id, item_rank)的结果集，按item_rank排序，然后在外层查询中使用collect_set。

查询

WITH table1 AS (
    SELECT 23 AS customer_id, 2 AS item_id, 3 AS item_rank UNION ALL
    SELECT 23 AS customer_id, 2 AS item_id, 3 AS item_rank UNION ALL
    SELECT 23 AS customer_id, 4 AS item_id, 2 AS item_rank UNION ALL
    SELECT 25 AS customer_id, 5 AS item_id, 1 AS item_rank UNION ALL
    SELECT 25 AS customer_id, 4 AS item_id, 2 AS item_rank
)
SELECT
    subquery.customer_id,
    collect_set(subquery.item_id) AS item_id_set
FROM (
    SELECT
        table1.customer_id,
        table1.item_id,
        table1.item_rank
    FROM table1
    DISTRIBUTE BY
        table1.customer_id
    SORT BY
        table1.customer_id,
        table1.item_rank
) subquery
GROUP BY
    subquery.customer_id
;

结果

    customer_id item_id_set
0   23  [4,2]
1   25  [5,4]

子查询使用 DISTRIBUTE BY 来保证特定 customer_id 的所有行都路由到同一个 reducer。然后使用 SORT BY 在每个 reducer 中按 customer_id 和 item_rank 排序。我希望这足以满足要求，因为我没有注意到对最终结果集进行总排序的要求。 (如果需要通过 customer_id 进行总排序，那么我认为查询必须使用 ORDER BY ，这会导致执行速度变慢。)

在内部， collect_set UDAF 使用 Java LinkedHashSet ，它是一个顺序保留集合，因此子查询中使用的相同排序顺序将在外部查询的集合中维护。这在 Hive 代码库中可见:

https://github.com/apache/hive/blob/release-2.0.0/ql/src/java/org/apache/hadoop/hive/ql/udf/generic/GenericUDAFMkCollectionEvaluator.java#L93