这可以使用一系列自连接来完成，以在将结果组合到 2 张地图之前找到同一类别中的其他房间。
代码
CREATE TABLE `table` AS
SELECT 1 AS customer, 'A' AS category, 'aa' AS room, 'd1' AS `date` UNION ALL
SELECT 1 AS customer, 'A' AS category, 'bb' AS room, 'd2' AS `date` UNION ALL
SELECT 1 AS customer, 'B' AS category, 'cc' AS room, 'd3' AS `date` UNION ALL
SELECT 1 AS customer, 'C' AS category, 'aa' AS room, 'd1' AS `date` UNION ALL
SELECT 1 AS customer, 'C' AS category, 'bb' AS room, 'd2' AS `date` UNION ALL
SELECT 2 AS customer, 'A' AS category, 'aa' AS room, 'd3' AS `date` UNION ALL
SELECT 2 AS customer, 'A' AS category, 'bb' AS room, 'd4' AS `date` UNION ALL
SELECT 2 AS customer, 'C' AS category, 'bb' AS room, 'd4' AS `date` UNION ALL
SELECT 2 AS customer, 'C' AS category, 'ee' AS room, 'd5' AS `date` UNION ALL
SELECT 3 AS customer, 'D' AS category, 'ee' AS room, 'd6' AS `date`
;
SELECT
customer_rooms.customer,
collect(customer_rooms.room, customer_rooms.date) AS map_customer_room_date,
collect(
COALESCE(customer_category_rooms.room, category_rooms.room),
COALESCE(customer_category_rooms.date, category_rooms.date)) AS map_category_room_date
FROM `table` AS customer_rooms
JOIN `table` AS category_rooms ON customer_rooms.category = category_rooms.category
LEFT OUTER JOIN `table` AS customer_category_rooms ON customer_rooms.customer = customer_category_rooms.customer
AND category_rooms.category = customer_category_rooms.category
AND category_rooms.room = customer_category_rooms.room
WHERE (
customer_rooms.customer = customer_category_rooms.customer AND
customer_rooms.category = customer_category_rooms.category AND
customer_rooms.room = customer_category_rooms.room AND
customer_rooms.date = customer_category_rooms.date
)
OR (
customer_category_rooms.customer IS NULL AND
customer_category_rooms.category IS NULL AND
customer_category_rooms.room IS NULL AND
customer_category_rooms.date IS NULL
)
GROUP BY
customer_rooms.customer
;
结果集
1 {"aa":"d1","bb":"d2","cc":"d3"} {"aa":"d1","bb":"d2","cc":"d3","ee":"d5"}
2 {"aa":"d3","bb":"d4","ee":"d5"} {"aa":"d3","bb":"d4","ee":"d5"}
3 {"ee":"d6"} {"ee":"d6"}
说明
FROM `table` AS customer_rooms
首先，从最初的table 中提取结果。我们将此关系命名为customer_rooms。正如您在问题中已经指出的那样，这足以构建 map_customer_room_date。
JOIN `table` AS category_rooms ON customer_rooms.category = category_rooms.category
第一个自联接标识与customer_rooms 行中明确提到的房间具有相同类别的所有房间。我们将此关系命名为category_rooms。
LEFT OUTER JOIN `table` AS customer_category_rooms ON customer_rooms.customer = customer_category_rooms.customer
AND category_rooms.category = customer_category_rooms.category
AND category_rooms.room = customer_category_rooms.room
第二个自加入获取我们在category_rooms 中标识的房间，并尝试查找该房间是否已由customer_rooms 中标识的客户持有。我们将此关系命名为customer_category_rooms。这是LEFT OUTER JOIN，因为我们想要保留之前连接的所有行。结果将是 1) customer_rooms 和 customer_category_rooms 的值相同，因为客户已经拥有这个房间，或者 2) customer_category_rooms 的值将全部是 NULL，因为客户没有持有这个房间，但它是同一类别中的一个房间。这种区别将变得很重要，以便我们可以保留客户的date（如果他们已经拥有房间）。
接下来，我们需要过滤。
WHERE (
customer_rooms.customer = customer_category_rooms.customer AND
customer_rooms.category = customer_category_rooms.category AND
customer_rooms.room = customer_category_rooms.room AND
customer_rooms.date = customer_category_rooms.date
)
这包括客户在原始table 中明确持有的房间。
OR (
customer_category_rooms.customer IS NULL AND
customer_category_rooms.category IS NULL AND
customer_category_rooms.room IS NULL AND
customer_category_rooms.date IS NULL
)
这包括不是客户持有但与客户持有的房间属于同一类别的房间。
collect(customer_rooms.room, customer_rooms.date) AS map_customer_room_date,
map_customer_room_date 可以通过从表中收集原始数据来构建，我们将其别名为customer_rooms。
collect(
COALESCE(customer_category_rooms.room, category_rooms.room),
COALESCE(customer_category_rooms.date, category_rooms.date)) AS map_category_room_date
map_category_room_date 大楼更复杂。如果客户明确持有房间，那么我们希望保留该date。但是，如果客户没有明确地持有房间，那么我们希望能够使用具有重叠类别的另一行中的 room 和 date。为此，我们使用 Hive COALESCE 函数选择第一个不是 NULL 的值。如果客户已经拥有房间（如 customer_category_rooms 中的非 NULL 值所示），那么我们将使用它。如果不是，那么我们将使用来自 category_rooms 的值。
请注意，如果同一类别/房间组合可以映射到多个 date 值，则仍可能存在一些歧义。如果这很重要，那么您可能需要投入更多的工作来根据某些业务规则（例如使用最快的date）选择正确的date，或者映射到多个date 值而不是单个值。如果有类似的额外要求，这应该会给你一个很好的起点。