BigQuery中重叠间隔的数量

Question

给定一个间隔表，我可以有效地查询每个间隔开始时的当前打开间隔数（包括当前间隔本身）吗？

例如，下表给出：

 
start_time end_time 
     1  10 
     2  5 
     3  4 
     5  6 
     7  11 
     19  20 

我想下面的输出：

 
start_time count 
     1  1 
     2  2 
     3  3 
     5  3 
     7  2 
     19  1 

在小数据集，我可以对自己加入该数据集解决这个问题：

WITH intervals AS (
    SELECT 1 AS start, 10 AS end UNION ALL 
    SELECT 2, 5 UNION ALL 
    SELECT 3, 4 UNION ALL 
    SELECT 5, 6 UNION ALL 
    SELECT 7, 11 UNION ALL 
    SELECT 19, 20 
) 
SELECT 
    a.start_time, 
    count(*) 
FROM 
    intervals a CROSS JOIN intervals b 
WHERE 
    a.start_time >= b.start_time AND 
    a.start_time <= b.end_time 
GROUP BY a.start_time 
ORDER BY a.start_time 

对于大型数据集，CROSS JOIN既不切实际也不必要，因为ny给出答案只取决于少数前面的区间（当按start_time排序时）。事实上，在我拥有的数据集中，它超时。有没有更好的方法来实现这一目标？

Answer 1

... CROSS JOIN既不切实际也不必要...
有没有更好的方法来实现这个目标？

请尝试下面的BigQuery标准SQL。没有加入参与

#standardSQL 
SELECT 
    start_time, 
    (SELECT COUNT(1) FROM UNNEST(ends) AS e WHERE e >= start_time) AS cnt 
FROM (
    SELECT 
    start_time, 
    ARRAY_AGG(end_time) OVER(ORDER BY start_time) AS ends 
    FROM intervals 
) 
-- ORDER BY start_time 

您可以测试/使用下面的例子发挥它从你的问题

#standardSQL 
WITH intervals AS (
    SELECT 1 AS start_time, 10 AS end_time UNION ALL 
    SELECT 2, 5 UNION ALL 
    SELECT 3, 4 UNION ALL 
    SELECT 5, 6 UNION ALL 
    SELECT 7, 11 UNION ALL 
    SELECT 19, 20 
) 
SELECT 
    start_time, 
    (SELECT COUNT(1) FROM UNNEST(ends) AS e WHERE e >= start_time) AS cnt 
FROM (
    SELECT 
    start_time, 
    ARRAY_AGG(end_time) OVER(ORDER BY start_time) AS ends 
    FROM intervals 
) 
-- ORDER BY start_time