在SQL中获取给定时间的最小值

我有以下SQL表：

start_time          end_time            value
2016-01-01 00:00:00 2016-01-01 08:59:59 1
2016-01-01 06:00:00 2016-01-01 14:59:59 2
2016-01-01 12:00:00 2016-01-01 17:59:59 1.5
2016-01-01 03:00:00 2016-01-01 17:59:59 3

我想把它转换成：

start_time          end_time            min_value
2016-01-01 00:00:00 2016-01-01 08:59:59 1
2016-01-01 09:00:00 2016-01-01 11:59:59 2
2016-01-01 12:00:00 2016-01-01 17:59:59 1.5

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其中，min_值是给定时间点的最小值。是否可以在SQL中执行此操作？

我不确定我是否理解预期输出与输入的关系，但如果您只想将最小值与不同的开始时间、结束时间对相关联，则可以执行以下操作，例如：

#standardSQL
WITH T AS (
  SELECT TIMESTAMP '2016-01-01 00:00:00' AS start_time,
    TIMESTAMP '2016-01-01 08:59:59' AS end_time, 1 AS value UNION ALL
  SELECT TIMESTAMP '2016-01-01 06:00:00',
    TIMESTAMP '2016-01-01 14:59:59', 2 UNION ALL
  SELECT TIMESTAMP '2016-01-01 12:00:00',
    TIMESTAMP '2016-01-01 17:59:59', 1.5 UNION ALL
  SELECT TIMESTAMP '2016-01-01 3:00:00',
    TIMESTAMP '2016-01-01 17:59:59', 3
)
SELECT
  start_time,
  end_time,
  MIN(value) AS min_value
FROM T
GROUP BY start_time, end_time;

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嗯。这似乎很难。我认为以下策略会奏效：

将数据分为两部分，即开始时间和结束时间。 对于每个开始时间，计算当时有效的最小值。 对于每个结束时间，计算从该时间开始生效的最小值。 使用间隙和孤岛方法重新组合 我只是不能100%肯定你能在BQ中做到这一点，因为它涉及到非等分联接。但是

with starts as (
      select start_time as time,
             (select min(t2.value)
              from t t2
              where t.start_time between t2.start_time and t2.end_time
             ) as value
      from t
     ),
     ends as (
      select end_time as time,
             (select min(t2.value)
              from t t2
              where t2.end_time > t.end_time and
                    t2.start_time <= t.end_time
             ) as value
      from t
     )
select value, min(time), max(time)
from (select time,
             row_number() over (order by time) as seqnum,
             row_number() over (partition by value order by time) as seqnum_v
      from ((select s.* from starts) union all
            (select e.* from ends)
           ) t
     ) t
group by value, (seqnum - seqnum_v);

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试试下面。我认为它完全符合你的要求 正如您所看到的，我在您的示例中又添加了一个条目，使其更辣：o

WITH YourTable AS (
SELECT TIMESTAMP '2016-01-01 00:00:00' AS start_time, TIMESTAMP '2016-01-01 08:59:59' AS end_time, 1 AS value UNION ALL
SELECT TIMESTAMP '2016-01-01 06:00:00' AS start_time, TIMESTAMP '2016-01-01 14:59:59' AS end_time, 2 AS value UNION ALL
SELECT TIMESTAMP '2016-01-01 12:00:00' AS start_time, TIMESTAMP '2016-01-01 17:59:59' AS end_time, 1.5 AS value UNION ALL
SELECT TIMESTAMP '2016-01-01 03:00:00' AS start_time, TIMESTAMP '2016-01-01 17:59:59' AS end_time, 3 AS value UNION ALL
SELECT TIMESTAMP '2016-01-01 12:30:00' AS start_time, TIMESTAMP '2016-01-01 12:40:59' AS end_time, 1 AS value 
), 
Intervals AS (
  SELECT iStart AS start_time, LEAD(iStart) OVER(ORDER BY iStart) AS end_time
  FROM (
    SELECT DISTINCT iStart FROM (
      SELECT start_time AS iStart FROM YourTable UNION ALL 
      SELECT end_time AS iStart FROM YourTable )
  )
),
Intervals_Mins AS (
  SELECT b.start_time, b.end_time, MIN(value) AS min_value
  FROM YourTable AS a
  JOIN Intervals AS b
  ON b.start_time BETWEEN a.start_time AND a.end_time
  AND b.end_time BETWEEN a.start_time AND a.end_time
  GROUP BY b.start_time, b.end_time
),
Intervals_Group AS (
  SELECT start_time, end_time, min_value, IFNULL(SUM(flag) OVER(PARTITION BY CAST(min_value AS STRING) ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING), 0) AS time_group 
  FROM (
    SELECT start_time, end_time, min_value, IF(end_time = LEAD(start_time) OVER(PARTITION BY CAST(min_value AS STRING) ORDER BY start_time), 0, 1) AS flag
    FROM Intervals_Mins
  )
)
SELECT MIN(start_time) AS start_time, MAX(end_time) AS end_time, min_value
FROM Intervals_Group
GROUP BY min_value, time_group
-- ORDER BY start_time

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这是一个非常困难的问题。@GordonLinoff-同意，这里的问题比通常的问题复杂一点：o