我有一个表id_vectors
,其中包含id
及其对应的coordinates
。每个coordinates
都是一个重复字段,其中包含512个元素。
我正在寻找所有这些向量之间的成对余弦相似性,例如如果我有三个ids
1,2和3,那么我正在寻找一个表,它们之间具有余弦相似性(基于使用512坐标的计算),如下所示:
id1 id2 similarity
1 2 0.5
1 3 0.1
2 3 0.99
现在我的表格中有424,970个唯一的ID
及其对应的512维坐标。这意味着基本上我需要创建大约(424970 * 424969/2)唯一的ID对,并计算它们的相似性。
我首先尝试使用reference from here进行以下查询:
#standardSQL
with pairwise as
(SELECT t1.id as id_1, t1.coords as coord1, t2.id as id_2, t2.coords as coord2
FROM `project.dataset.id_vectors` t1
inner join `project.dataset.id_vectors` t2
on t1.id < t2.id)
SELECT id_1, id_2, (
SELECT
SUM(value1 * value2)/
SQRT(SUM(value1 * value1))/
SQRT(SUM(value2 * value2))
FROM UNNEST(coord1) value1 WITH OFFSET pos1
JOIN UNNEST(coord2) value2 WITH OFFSET pos2
ON pos1 = pos2
) cosine_similarity
FROM pairwise
但是运行6个小时后,我遇到了以下错误消息
Query exceeded resource limits. 2.2127481953201417E7 CPU seconds were used, and this query must use less than 428000.0 CPU seconds.
然后,我想而不是使用中间表pairwise
,为什么不先尝试创建该表,然后再进行余弦相似度计算。
所以我尝试了以下查询:
SELECT t1.id as id_1, t1.coords as coord1, t2.id as id_2, t2.coords as coord2
FROM `project.dataset.id_vectors` t1
inner join `project.dataset.id_vectors` t2
on t1.id < t2.id
但是这次查询无法完成,我遇到了以下消息:
Error: Quota exceeded: Your project exceeded quota for total shuffle size limit. For more information, see https://cloud.google.com/bigquery/troubleshooting-errors
。
然后,我尝试通过创建ID的组合对并使用以下查询从其剥离坐标来创建甚至更小的表:
SELECT t1.id as id_1, t2.id as id_2
FROM `project.dataset.id_vectors` t1
inner join `project.dataset.id_vectors` t2
on t1.id < t2.id
我的查询再次以错误消息Query exceeded resource limits. 610104.3843576935 CPU seconds were used, and this query must use less than 3000.0 CPU seconds. (error code: billingTierLimitExceeded)
我完全理解这是一个巨大的查询,而我的出发点是我的帐单配额。
我要问的是,有没有一种方法可以以更智能的方式执行查询,以使我不超过resourceLimit
,shuffleSizeLimit
或billingTierLimit
中的任何一个?< / p>
答案 0 :(得分:1)
一个简单的想法是-与其将表与多余的坐标连接起来,而是应该创建简单的对表(id1,id2),然后通过两个额外的连接来“修饰”各自的id及其坐标向量到dataset.table.id_vectors
下面是一个简短的示例:
#standardSQL
WITH pairwise AS (
SELECT t1.id AS id_1, t2.id AS id_2
FROM `project.dataset.id_vectors` t1
INNER JOIN `project.dataset.id_vectors` t2
ON t1.id < t2.id
)
SELECT id_1, id_2, (
SELECT
SUM(value1 * value2)/
SQRT(SUM(value1 * value1))/
SQRT(SUM(value2 * value2))
FROM UNNEST(a.coords) value1 WITH OFFSET pos1
JOIN UNNEST(b.coords) value2 WITH OFFSET pos2
ON pos1 = pos2
) cosine_similarity
FROM pairwise t
JOIN `project.dataset.id_vectors` a ON a.id = id_1
JOIN `project.dataset.id_vectors` b ON b.id = id_2
很明显,它适用于小型虚拟集,如下所示:
#standardSQL
WITH `project.dataset.id_vectors` AS (
SELECT 1 id, [1.0, 2.0, 3.0, 4.0] coords UNION ALL
SELECT 2, [1.0, 2.0, 3.0, 4.0] UNION ALL
SELECT 3, [2.0, 0.0, 1.0, 1.0] UNION ALL
SELECT 4, [0, 2.0, 1.0, 1.0] UNION ALL
SELECT 5, [2.0, 1.0, 1.0, 0.0] UNION ALL
SELECT 6, [1.0, 1.0, 1.0, 1.0]
), pairwise AS (
SELECT t1.id AS id_1, t2.id AS id_2
FROM `project.dataset.id_vectors` t1
INNER JOIN `project.dataset.id_vectors` t2
ON t1.id < t2.id
)
SELECT id_1, id_2, (
SELECT
SUM(value1 * value2)/
SQRT(SUM(value1 * value1))/
SQRT(SUM(value2 * value2))
FROM UNNEST(a.coords) value1 WITH OFFSET pos1
JOIN UNNEST(b.coords) value2 WITH OFFSET pos2
ON pos1 = pos2
) cosine_similarity
FROM pairwise t
JOIN `project.dataset.id_vectors` a ON a.id = id_1
JOIN `project.dataset.id_vectors` b ON b.id = id_2
有结果
Row id_1 id_2 cosine_similarity
1 1 2 1.0
2 1 3 0.6708203932499369
3 1 4 0.819891591749923
4 1 5 0.521749194749951
5 1 6 0.9128709291752769
6 2 3 0.6708203932499369
7 2 4 0.819891591749923
8 2 5 0.521749194749951
9 2 6 0.9128709291752769
10 3 4 0.3333333333333334
11 3 5 0.8333333333333335
12 3 6 0.8164965809277261
13 4 5 0.5000000000000001
14 4 6 0.8164965809277261
15 5 6 0.8164965809277261
因此,请尝试使用您的真实数据,让我们看看它如何为您工作:o)
而且...显然,您应该预先创建/实现pairwise
表
另一个优化想法是在您的SQRT(SUM(value1 * value1))
中预先计算project.dataset.id_vectors
的值-这样可以节省大量CPU-这应该是简单的调整,所以我留给您:o)