根据事件发生的时间分配唯一的组 ID（彼此相隔 60 分钟以内）

Question

有没有一种方法可以根据每行之间的时间间隔（在本例中为 60 分钟）为 SQL 中的行分配唯一的组 ID。本质上，我需要它从第一行开始（8 月 27 日下午 1 点），这将是第 1 行，然后开始遍历每一行（看看是否有一行是 <= 60 min from 1pm), if it is it would be #2, then #3, etc. But once its over 60 min, I then need it to start over again at #1, and use the same logic, going through each row to see if <= 60 min. I've included a screen shot of the expected Groupid ID I would like to see.

尝试使用 LEAD/LAG 的某种组合但不成功

Answer 1

这听起来像是经典的“重叠日期范围”的一部分，如果您想了解更多详细信息，可以通过谷歌搜索。另一个你会发现很多文献的类似问题是“差距和孤岛问题”。

基本上，您想执行以下步骤：

1. 添加一个新列 + 60 分钟，或者任何你的缓冲区，以获得“结束时间”
2. 超前和滞后，这样你就可以计算与“下一个”的差异，以及计算一个岛屿是否“开始”

我有一个非常通用的template，我为这个问题运行，它可能包含很多你不需要的东西。它还适用于 Snowflake 风格，因此您可能必须针对您的特定 RDBMS 对其进行调整。但希望这对您有所帮助：

WITH CTE_CONDITION AS (
  SELECT 
    BP_DATETIME AS dtm 
  FROM 
    ExampleTable 
  WHERE 
    FIRST_BP_READING IS NOT NULL 
    AND BP_DATETIME is not null
), 
CTE_LAGGED AS (
  SELECT 
    dtm, 
    LAG(dtm) OVER (
      ORDER BY 
        dtm
    ) AS previous_datetime, 
    LEAD(dtm) OVER (
      ORDER BY 
        dtm
    ) AS next_datetime, 
    ROW_NUMBER() OVER (
      ORDER BY 
        CTE_CONDITION.dtm
    ) AS island_location 
  FROM 
    CTE_CONDITION
), 
CTE_ISLAND_START AS (
  SELECT 
    ROW_NUMBER() OVER (
      ORDER BY 
        dtm
    ) AS island_number, 
    dtm AS island_start_datetime, 
    island_location AS island_start_location 
  FROM 
    CTE_LAGGED 
  WHERE 
    (
      DATEDIFF(HOUR, previous_datetime, dtm) > 60 
      OR CTE_LAGGED.previous_datetime IS NULL
    )
), 
CTE_ISLAND_END AS (
  SELECT 
    ROW_NUMBER() OVER (
      ORDER BY 
        dtm
    ) AS island_number, 
    dtm AS island_end_datetime, 
    island_location AS island_end_location 
  FROM 
    CTE_LAGGED 
  WHERE 
    DATEDIFF(HOUR, dtm, next_datetime) > 60 
    OR CTE_LAGGED.next_datetime IS NULL
) 
SELECT 
  CTE_ISLAND_START.island_start_datetime, 
  CTE_ISLAND_END.island_end_datetime, 
  DATEDIFF(
    HOUR, CTE_ISLAND_START.island_start_datetime, 
    CTE_ISLAND_END.island_end_datetime
  ) AS ISLAND_DURATION_HOUR, 
  (
    SELECT 
      COUNT(*) 
    FROM 
      CTE_LAGGED 
    WHERE 
      CTE_LAGGED.dtm BETWEEN CTE_ISLAND_START.island_start_datetime 
      AND CTE_ISLAND_END.island_end_datetime
  ) AS island_row_count 
FROM 
  CTE_ISLAND_START 
  INNER JOIN CTE_ISLAND_END ON CTE_ISLAND_END.island_number = CTE_ISLAND_START.island_number

Answer 2

Postgres

根据时间间隔为 SQL 中的行分配唯一组 ID 的一种方法是使用带窗口函数的递归 CTE（通用表表达式）。这是一个应该达到预期结果的示例查询：

WITH RECURSIVE t AS (
  SELECT timestamp, 
         ROW_NUMBER() OVER (ORDER BY timestamp) AS row_num
  FROM your_table
), rcte AS (
  SELECT row_num, 
         timestamp, 
         1 AS group_id
  FROM t
  WHERE row_num = 1
  UNION ALL
  SELECT t.row_num, 
         t.timestamp, 
         CASE WHEN EXTRACT(epoch FROM (t.timestamp - rcte.timestamp)) <= 3600 
              THEN rcte.group_id 
              ELSE rcte.group_id + 1 
         END AS group_id
  FROM t
  JOIN rcte ON t.row_num = rcte.row_num + 1
)
SELECT row_num, group_id
FROM rcte
ORDER BY row_num;

这是查询的作用：

1. t 子查询从 your_table 中选择时间戳列，并使用 ROW_NUMBER() 窗口函数为每一行分配一个行号。
2. rcte CTE 是查询的递归部分。它从第一行开始（其中 row_num 为 1）并为其分配组 ID 1。然后，对于每个后续行，它检查当前行的时间戳与前一行的时间戳之间的秒数差异是否小于或等于到 3600（60 分钟）。如果是，它会分配与前一行相同的组 ID。如果不是，它会通过将前一行的组 ID 递增 1 来分配一个新的组 ID。
3. 最后，主查询从 rcte CTE 中选择 row_num 和 group_id 列，并按 row_num 对结果进行排序。

请注意，此查询假定时间戳列已按升序排序。如果不是，您可以将 ORDER BY 子句添加到 t 子查询以确保行被正确排序。

微软 SQL 服务器

Microsoft SQL Server 不支持

WITH RECURSIVE

语法。相反，您可以使用公用表表达式 (CTE) 和带有 UNION ALL 运算符的递归查询来获得相同的结果。这是查询的示例转换：

WITH t AS (
  SELECT timestamp, 
         ROW_NUMBER() OVER (ORDER BY timestamp) AS row_num
  FROM your_table
), rcte AS (
  SELECT row_num, 
         timestamp, 
         1 AS group_id
  FROM t
  WHERE row_num = 1
  UNION ALL
  SELECT t.row_num, 
         t.timestamp, 
         CASE WHEN DATEDIFF(second, t.timestamp, rcte.timestamp) <= 3600 
              THEN rcte.group_id 
              ELSE rcte.group_id + 1 
         END AS group_id
  FROM t
  JOIN rcte ON t.row_num = rcte.row_num + 1
)
SELECT row_num, group_id
FROM rcte
ORDER BY row_num;

注意，我用

EXTRACT(epoch FROM ...)

函数替换了

DATEDIFF(second, ...)

函数，它以秒为单位计算时差。