Question

我正在尝试解决一个问题，其中我想合并给定列ID的重叠间隔，但我也想跟踪每个重叠间隔的最大值。每个间隔都有start_time和stop_time，每个间隔都有一个与之关联的层次结构/优先级。

这些是表格中的以下列： id，start_time，stop_time，some_value

示例输入：

示例输出：

Answer 1

以下是用于BigQuery标准SQL的代码，我假设您仍在处理与上一个问题相同的用例，因此我想使其与该解决方案保持一致-可以在需要考虑的情况下对其进行扩展优先级

所以，无论如何：

#standardSQL
WITH check_times AS (
  SELECT id, start_time AS TIME FROM `project.dataset.table` UNION DISTINCT
  SELECT id, stop_time AS TIME FROM `project.dataset.table` 
), distinct_intervals AS (
  SELECT id, TIME AS start_time, LEAD(TIME) OVER(PARTITION BY id ORDER BY TIME) stop_time
  FROM check_times
), deduped_intervals AS (
  SELECT a.id, a.start_time, a.stop_time, MAX(some_value) some_value 
  FROM distinct_intervals a
  JOIN `project.dataset.table` b
  ON a.id = b.id 
  AND a.start_time BETWEEN b.start_time AND b.stop_time 
  AND a.stop_time BETWEEN b.start_time AND b.stop_time
  GROUP BY a.id, a.start_time, a.stop_time
), combined_intervals AS (
  SELECT id, MIN(start_time) start_time, MAX(stop_time) stop_time, MAX(some_value) some_value 
  FROM (
    SELECT id, start_time, stop_time, some_value, COUNTIF(flag) OVER(PARTITION BY id ORDER BY start_time) grp
    FROM (
      SELECT id, start_time, stop_time, some_value,
        start_time != IFNULL(LAG(stop_time) OVER(PARTITION BY id ORDER BY start_time), start_time) flag
      FROM deduped_intervals
    )
  )
  GROUP BY id, grp
)
SELECT *
FROM combined_intervals
-- ORDER BY id, start_time

如果要应用于您的样本数据-结果为

Row id  start_time  stop_time   some_value   
1   1   0           36          50   
2   1   41          47          23

是否可以在结果中再增加一列，以显示该时间段内的事件数

#standardSQL
WITH check_times AS (
  SELECT id, start_time AS TIME FROM `project.dataset.table` UNION DISTINCT
  SELECT id, stop_time AS TIME FROM `project.dataset.table` 
), distinct_intervals AS (
  SELECT id, TIME AS start_time, LEAD(TIME) OVER(PARTITION BY id ORDER BY TIME) stop_time
  FROM check_times
), deduped_intervals AS (
  SELECT a.id, a.start_time, a.stop_time, MAX(some_value) some_value, ANY_VALUE(To_JSON_STRING(b)) event_hash
  FROM distinct_intervals a
  JOIN `project.dataset.table` b
  ON a.id = b.id 
  AND a.start_time BETWEEN b.start_time AND b.stop_time 
  AND a.stop_time BETWEEN b.start_time AND b.stop_time
  GROUP BY a.id, a.start_time, a.stop_time
), combined_intervals AS (
  SELECT id, MIN(start_time) start_time, MAX(stop_time) stop_time, MAX(some_value) some_value, COUNT(DISTINCT event_hash) events
  FROM (
    SELECT *, COUNTIF(flag) OVER(PARTITION BY id ORDER BY start_time) grp
    FROM (
      SELECT *,
        start_time != IFNULL(LAG(stop_time) OVER(PARTITION BY id ORDER BY start_time), start_time) flag
      FROM deduped_intervals
    )
  )
  GROUP BY id, grp
)
SELECT *
FROM combined_intervals
-- ORDER BY id, start_time

有结果

Row id  start_time  stop_time   some_value  events   
1   1   0           36          50          8    
2   1   41          47          23          1

Answer 2

您可以使用累积的max()确定何时开始新分组。然后用累积条件count()来识别组。。。最后聚合：

select min(start_time), max(stop_time), max(some_value)
from (select t.*,
             countif(prev_stop_time is null or prev_stop_time < start_time) over (partition by id order by start_time) as grp
      from (select t.*,
                   max(stop_time) over (partition by id order by start_time rows between unbounded preceding and 1 preceding) as prev_stop_time
            from t
           ) t
     ) t
group by item_id, grp;

合并重叠间隔并跟踪BigQuery SQL中的最大值

2 个答案: