根据应用于 BigQuery 中另一个嵌套列的条件从嵌套列中选择值

Question

如何使用嵌套列中“特殊”值的索引（例如：该嵌套列中最大值的索引）select使用该索引的另一个嵌套列中的值？

例如，考虑具有以下架构的 table：

Field name	Type	Mode
id	STRING	NULLABLE
username	STRING	NULLABLE
▼ products	RECORD	NULLABLE
▼ list	RECORD	REPEATED
item	STRING	NULLABLE
▼ ordered	RECORD	NULLABLE
▼ list	RECORD	REPEATED
item	INTEGER	NULLABLE
total_orders	STRING	NULLABLE
update_time	TIMESTAMP	NULLABLE
update_id	INTEGER	NULLABLE

前几行如下所示：

Row	id	username	products.list.item	ordered.list.item	total_orders	update_time	update_id
1	1234	a_turing	Apple	1	3	2021-08-14 20:03:22.100846 UTC	121231
			Orange	0
			Pear	2
2	5678	g_hopper	Apple	0	2	2021-08-15 09:36:48.220464 UTC	121232
			Orange	2
			Pear	0
3	1122	a_lovelace	Apple	0	1	2021-08-15 13:59:03.441506 UTC	121233
			Orange	1
			Pear	0
4	3344	v_nabokov	Apple	1	2	2021-08-17 17:34:53.415406 UTC	121234
			Orange	0
			Pear	1

我想 select 每个 id 的最近订单的最常订购产品，并排除没有最常订购产品的订单（例如，如果客户订购了相同数量的 Apple、Orange 和梨).

我目前使用的查询是一个 CTE 链，每个产品类型一个，外加一个额外的列，即每个用户订购的产品的最大数量 (max_ordered)。然后我使用 id 列将 CTE 连接在一起：

WITH RANKED_ORDERS AS( 
SELECT *, ROW_NUMBER() OVER (PARTITION BY id ORDER BY update_time DESC) AS rn
FROM mycompany.engagement.products_ordered),

LATEST_ORDERS AS(
SELECT * FROM RANKED_ORDERS WHERE rn = 1),

-- ---------------------- Apples Ordered -----------------------
APPLES_INDEXED AS(
SELECT id, username, ordered, flattened_products, offset_nk, update_time, rn
FROM LATEST_ORDERS
CROSS JOIN UNNEST(LATEST_ORDERS.products.list) AS flattened_products
WITH OFFSET as offset_nk
WHERE flattened_products.item in ('Apple')
ORDER BY offset_nk),

APPLES_ORDERED AS(
SELECT id, username, update_time, ordered.list[OFFSET(offset_nk)].item as apples_ordered 
FROM APPLES_INDEXED 
ORDER BY
update_time ASC),

-- ---------------------- Oranges Ordered ----------------------
ORANGES_INDEXED AS(
SELECT id, username, ordered, flattened_products, offset_nk, update_time, rn
FROM LATEST_ORDERS
CROSS JOIN UNNEST(LATEST_ORDERS.products.list) AS flattened_products
WITH OFFSET as offset_nk
WHERE flattened_products.item in ('Orange')
ORDER BY offset_nk),

ORANGES_ORDERED AS(
SELECT id, username, update_time, ordered.list[OFFSET(offset_nk)].item as oranges_ordered 
FROM ORANGES_INDEXED 
ORDER BY
update_time ASC),

-- ---------------------- Pears Ordered -----------------------
PEARS_INDEXED AS(
SELECT id, username, ordered, flattened_products, offset_nk, update_time, rn
FROM LATEST_ORDERS
CROSS JOIN UNNEST(LATEST_ORDERS.products.list) AS flattened_products
WITH OFFSET as offset_nk
WHERE flattened_products.item in ('Pear')
ORDER BY offset_nk),

PEARS_ORDERED AS(
SELECT id, username, update_time, ordered.list[OFFSET(offset_nk)].item as pears_ordered 
FROM PEARS_INDEXED 
ORDER BY
update_time ASC),

-- --------------- Max Product Ordered per Order --------------
MAX_ORDERED AS(
SELECT
id, username, MAX(orders_per_username.item) as max_ordered, total_orders
FROM
LATEST_ORDERS, UNNEST(ordered.list) as orders_per_username
GROUP BY id, username, total_orders),

-- -------------------- Orders In Columns ---------------------
ORDERS_IN_COLUMNS AS(
SELECT APPLES_ORDERED.username, APPLES_ORDERED.update_time, APPLES_ORDERED.apples_ordered,
ORANGES_ORDERED.oranges_ordered, PEARS_ORDERED.pears_ordered, MAX_ORDERED.max_ordered
FROM APPLES_ORDERED
LEFT JOIN ORANGES_ORDERED ON ORANGES_ORDERED.id = APPLES_ORDERED.id
LEFT JOIN PEARS_ORDERED ON PEARS_ORDERED.id = APPLES_ORDERED.id
LEFT JOIN MAX_ORDERED ON MAX_ORDERED.id = APPLES_ORDERED.id),

-- ------- Orders with a most ordered product -----------------
NO_CONFLICTS AS(
SELECT * FROM ORDERS_IN_COLUMNS
WHERE
max_ordered > 0 AND
(
    (apples_ordered not in (oranges_ordered, pears_ordered) AND apples_ordered = max_ordered)
OR
    (oranges_ordered not in (apples_ordered, pears_ordered) AND oranges_ordered = max_ordered)
OR
    (pears_ordered not in (apples_ordered, oranges_ordered) AND pears_ordered = max_ordered)
)
)

SELECT * FROM NO_CONFLICTS

此 return 是以下 table：

Row	username	update_time	apples_ordered	oranges_ordered	pears_ordered	max_ordered
1	a_turing	2021-08-14 20:03:22.100846 UTC	1	0	2	2
2	g_hopper	2021-08-15 09:36:48.220464 UTC	0	2	0	2
3	a_lovelace	2021-08-15 13:59:03.441506 UTC	0	1	0	1

这是意料之中的。
但是，我无法弄清楚如何简单地 return a table 看起来像：

Row	username	update_time	max_product_ordered
1	a_turing	2021-08-14 20:03:22.100846 UTC	Pear
2	g_hopper	2021-08-15 09:36:48.220464 UTC	Orange
3	a_lovelace	2021-08-15 13:59:03.441506 UTC	Orange

我也相当确定，虽然这个查询基本上有效（我最终在 Python 中进行 post 处理以到达最后一步）它可能鉴于“常见 table 表达式”的广泛使用，效率极低。
是否有比我编写的方法更有效的查询 BigQuery table 的方法，或者我是否需要完全重组 table 以获得任何加速？目前，运行这个查询需要大约 10 秒才能完成 table 上的查询，大约有 10,000 行和 12 列，我认为这种缓慢是由于多个 CTE。
在过去的两周里，我一直在用头撞墙，试图改进我的查询，但没有取得太大进展。 真诚地感谢任何帮助！

Answer 1

考虑以下方法

with latest_orders as (
  select * from `mycompany.engagement.products_ordered`
  where true 
  qualify 1 = row_number() over(partition by id order by update_time desc)
), qualified_items as (
  select *, 
    array(
      select offset from t.ordered.list with offset 
      where true 
      qualify 1 = rank() over(order by item desc) 
    ) items
  from latest_orders t
)
select id, username, update_time,
  products.list[offset(items[offset(0)])] as max_product_ordered,
from qualified_items
where array_length(items) = 1

如果应用于您问题中的示例数据 - 输出为

根据应用于 BigQuery 中另一个嵌套列的条件从嵌套列中选择值

Selecting values from a nested column based on a condition applied to another nested column in BigQuery

sql

common-table-expression

google-bigquery