如何在 BigQuery 中获取连续时间戳之间的差异

Posted 2023-03-24

【中文标题】如何在 BigQuery 中获取连续时间戳之间的差异

【英文标题】：How to Get Difference Between Consecutive TimeStamps in BigQuery

【发布时间】：2016-07-26 19:39:51

【问题描述】：

我有一张如下所示的表格：

ID    DateTime
1     5-1-16 12:25:13
1     5-1-16 12:28:46
2     5-1-16 12:25:18
2     5-1-16 12:29:34

我想找出每个 ID 的每个连续时间戳之间的秒数差异。有没有办法在 BigQuery 中做到这一点？我有几千条记录。我知道我需要先将时间与日期隔离开来。

【参考方案1】：

下面试试

SELECT
  ID, 
  TIMESTAMP_TO_SEC(TIMESTAMP(DateTime))-TIMESTAMP_TO_SEC(TIMESTAMP(prev_DateTime)) AS diff,
FROM (
  SELECT 
    ID,
    DateTime,
    LAG(DateTime) OVER(PARTITION BY ID ORDER BY DateTime) AS prev_DateTime
  FROM 
    (SELECT 1 AS ID, '2016-05-01 12:25:13' AS DateTime),
    (SELECT 1 AS ID, '2016-05-01 12:28:46' AS DateTime),
    (SELECT 2 AS ID, '2016-05-01 12:25:18' AS DateTime),
    (SELECT 2 AS ID, '2016-05-01 12:29:34' AS DateTime)
)

添加

 SELECT
  ID, 
  TIMESTAMP_TO_SEC(TIMESTAMP(DateTime))-TIMESTAMP_TO_SEC(TIMESTAMP(prev_DateTime)) AS diff,
FROM (
  SELECT 
    ID,
    DateTime,
    LAG(DateTime) OVER(PARTITION BY ID ORDER BY DateTime) AS prev_DateTime
  FROM YourTable
)

我注意到 - 您的 DateTime 字段的格式似乎不寻常 - '5-1-16 12:29:34' 如果在实施上述查询时这对您来说是个问题 - 您可以尝试以下一个 请注意：对于此查询，您需要enable Standard SQL

SELECT
  ID, 
  UNIX_SECONDS(DateTime) - UNIX_SECONDS(prev_DateTime) AS diff
FROM (
  SELECT 
    ID,
    DateTime,
    LAG(DateTime) OVER(PARTITION BY ID ORDER BY DateTime) AS prev_DateTime
  FROM (
    SELECT
      ID, 
      PARSE_TIMESTAMP("%m-%d-%y %H:%M:%S", DateTime) AS DateTime
    FROM YourTable  
  )
)

【讨论】：

感谢您的回复。但是，我有几千个 ID 和日期时间，如何在不输入每个 ID 和日期时间的情况下修改此查询？

在 FROM 后面使用你的表名。这几个选择被用作你的例子:o) 见补充回答