BigQuery 根据最接近的时间戳和匹配值组合表答案

【问题标题】：BigQuery combine tables based on closest timerstamp and matching valueBigQuery 根据最接近的时间戳和匹配值组合表
【发布时间】：2016-11-03 23:45:14
【问题描述】：

我有两个表，对于表 numberTwo 的每一行，我需要在表 numberOne 中获取具有相同 的 hint >cod 值以及在比较 time1 和 time2 时具有 最接近时间 的值。

为了更容易理解我需要做的是：

表号一：

|  id |  cod  |   hint  |           time1         |
---------------------------------------------------
|  1  |  ABC  |    V    | 2016-11-03 18:00:00 UTC |
|  2  |  ABC  |    W    | 2016-11-03 12:00:00 UTC |
|  3  |  CDE  |    X    | 2016-11-03 19:00:00 UTC |
|  4  |  CDE  |    Y    | 2016-11-03 19:30:00 UTC |
|  5  |  EFG  |    Z    | 2016-11-03 18:00:00 UTC |

表号二

|  id |  cod  |   value  |         time2           |
----------------------------------------------------
|  1  |  ABC  |   xyz2   | 2016-11-03 18:20:00 UTC |
|  2  |  ABC  |   h323   | 2016-11-03 11:30:00 UTC |
|  3  |  ABC  |   rewq   | 2016-11-03 09:00:00 UTC |
|  4  |  CDE  |   abce   | 2016-11-03 19:10:00 UTC |

因此，对于表 numberTwo 的 row #1，我将使用 cod: ABC 获取表 numberOne 中的所有行强>

|  1  |  ABC  |    V    | 2016-11-03 18:00:00 UTC |
|  2  |  ABC  |    W    | 2016-11-03 12:00:00 UTC |

在这两者之间，我会得到一个与 time2 最接近的时间戳：

|  1  |  ABC  |    V    | 2016-11-03 18:00:00 UTC |

处理完每一行后，我会得到一个像这样的表格：

所需的表

|  id |  cod  |   hint  |   value  |         time2           |
--------------------------------------------------------------
|  1  |  ABC  |    V    |   xyz2   | 2016-11-03 18:20:00 UTC |
|  2  |  ABC  |    W    |   h323   | 2016-11-03 11:30:00 UTC |
|  3  |  ABC  |    W    |   rewq   | 2016-11-03 09:00:00 UTC |
|  4  |  CDE  |    X    |   abce   | 2016-11-03 19:10:00 UTC |

【问题讨论】：

标签： mysql sql google-bigquery

【解决方案1】：

对于 BigQuery 标准 SQL - 在下面尝试

您可以取消注释带有示例数据的注释块以进行快速测试

WITH 
/*    
TableNumberOne AS (
  SELECT 1 AS id, 'ABC' AS cod, 'V' AS hint, TIMESTAMP '2016-11-03 18:00:00 UTC' AS time1 UNION ALL
  SELECT 2 AS id, 'ABC' AS cod, 'W' AS hint, TIMESTAMP '2016-11-03 12:00:00 UTC' AS time1 UNION ALL
  SELECT 3 AS id, 'CDE' AS cod, 'X' AS hint, TIMESTAMP '2016-11-03 19:00:00 UTC' AS time1 UNION ALL
  SELECT 4 AS id, 'CDE' AS cod, 'Y' AS hint, TIMESTAMP '2016-11-03 19:30:00 UTC' AS time1 UNION ALL
  SELECT 5 AS id, 'EFG' AS cod, 'Z' AS hint, TIMESTAMP '2016-11-03 18:00:00 UTC' AS time1 
),
TableNumberTwo AS (
  SELECT 1 AS id, 'ABC' AS cod, 'xyz2' AS value, TIMESTAMP '2016-11-03 18:20:00 UTC' AS time2 UNION ALL
  SELECT 2 AS id, 'ABC' AS cod, 'h323' AS value, TIMESTAMP '2016-11-03 11:30:00 UTC' AS time2 UNION ALL
  SELECT 3 AS id, 'ABC' AS cod, 'rewq' AS value, TIMESTAMP '2016-11-03 09:00:00 UTC' AS time2 UNION ALL
  SELECT 4 AS id, 'CDE' AS cod, 'abce' AS value, TIMESTAMP '2016-11-03 19:10:00 UTC' AS time2 
),
*/
tempTable AS (
  SELECT 
    t2.id, t2.cod, t2.value, t2.time2, t1.hint, 
    ROW_NUMBER() OVER(PARTITION BY t2.id, t2.cod, t2.value 
                      ORDER BY ABS(TIMESTAMP_DIFF(t2.time2, t1.time1, SECOND))) AS win
  FROM TableNumberTwo AS t2
  JOIN TableNumberOne AS t1
  ON t1.cod = t2.cod
)
SELECT id, cod, hint, value, time2
FROM tempTable
WHERE win = 1

【讨论】：

我是否应该保留不匹配的行，我该怎么做？（那些提示将是 NULL ......）
你会使用 LEFT JOIN 而不是 JOIN
哦，好吧，现在我开始明白了哈哈。非常感谢你的帮助，你是最棒的。
还有其他方法吗？因为如果我使用左连接（包括在另一个问题中），从 68 开始的计费层基本上是无限的（需要 4628414464 或更高。）并且不断上升，使得无法运行查询。
哇，不好。即使是68也足够大。我会想。今晚晚些时候回来。

【解决方案2】：

还有其他方法吗？因为如果我使用左连接（包括在其他问题）从 68 开始的计费等级基本上是无限的（需要 4628414464 或更高版本。）并且不断上升无法运行查询

我觉得可以玩的地方很少

1 - ABS(TIMESTAMP_DIFF(t2.time2, t1.time1, SECOND)) - 此函数针对连接中的所有排列运行。您可能想尝试在单独的子选择中将每个表的相应时间字段转换为秒，而不是使用它而不是原始表 - 因此您只需要执行类似 ABS(t2.time2inSeconds - t1.time1inSeconds) 的操作

2 - 使用ROW_NUMBER() 是另一个潜在的费用来源 - 请参阅下面的查询，我试图完全重写逻辑 - 但这是非常盲目的尝试，因为我无法对其进行测试，看看这是否真的修复或改进了东西或不。如果您可以尝试并告知结果（计费等级），那就太好了

WITH 
/*    
TableNumberOne AS (
  SELECT 1 AS id, 'ABC' AS cod, 'V' AS hint, TIMESTAMP '2016-11-03 18:00:00 UTC' AS time1 UNION ALL
  SELECT 2 AS id, 'ABC' AS cod, 'W' AS hint, TIMESTAMP '2016-11-03 12:00:00 UTC' AS time1 UNION ALL
  SELECT 3 AS id, 'CDE' AS cod, 'X' AS hint, TIMESTAMP '2016-11-03 19:00:00 UTC' AS time1 UNION ALL
  SELECT 4 AS id, 'CDE' AS cod, 'Y' AS hint, TIMESTAMP '2016-11-03 19:30:00 UTC' AS time1 UNION ALL
  SELECT 5 AS id, 'EFG' AS cod, 'Z' AS hint, TIMESTAMP '2016-11-03 18:00:00 UTC' AS time1 
),
TableNumberTwo AS (
  SELECT 1 AS id, 'ABC' AS cod, 'xyz2' AS value, TIMESTAMP '2016-11-03 18:20:00 UTC' AS time2 UNION ALL
  SELECT 2 AS id, 'ABC' AS cod, 'h323' AS value, TIMESTAMP '2016-11-03 11:30:00 UTC' AS time2 UNION ALL
  SELECT 3 AS id, 'ABC' AS cod, 'rewq' AS value, TIMESTAMP '2016-11-03 09:00:00 UTC' AS time2 UNION ALL
  SELECT 4 AS id, 'CDE' AS cod, 'abce' AS value, TIMESTAMP '2016-11-03 19:10:00 UTC' AS time2 
),
*/
tempTable1 AS (
  SELECT 
    t2.id, t2.cod, t2.value, 
    MIN(ABS(TIMESTAMP_DIFF(t2.time2, t1.time1, SECOND))) AS delta 
  FROM TableNumberTwo AS t2
  JOIN TableNumberOne AS t1
  ON t1.cod = t2.cod
  GROUP BY t2.id, t2.cod, t2.value
),
tempTable2 AS (
  SELECT a.id, a.cod, a.value, a.time2, b.delta
  FROM TableNumberTwo AS a 
  JOIN tempTable1 AS b 
  ON a.id = b.id AND a.cod = b.cod AND a.value = b.value
)
SELECT a.id, a.cod, t1.hint, a.value, a.time2
FROM tempTable2 AS a
JOIN TableNumberOne AS t1
ON t1.cod = a.cod AND ABS(TIMESTAMP_DIFF(a.time2, t1.time1, SECOND)) = delta

注意：上面的查询仍然应该是完整的，因为它可以从 tableOne 返回几个匹配的行，这些行与 tableTwo 中的相应行同样接近。但就目前而言 - 只是为了验证成本问题是否已得到修复或至少得到改进

3 - 顺便说一句，它也可以是你的倾斜数据等。

【讨论】：