COUNT() OVER 以当前行为条件

Question

给定代表一个任务的每一行，以及开始时间和结束时间，我如何使用窗口函数计算每个任务开始（包括其本身）时正在运行的任务（即开始和未结束）的数量COUNT OVER?窗口函数是正确的方法吗？

例如，给定表tasks：

task_id  start_time  end_time
   a         1          10
   b         2           5
   c         5          15
   d         8          13
   e        12          20
   f        21          30

计算running_tasks:

task_id  start_time  end_time  running_tasks
   a         1          10           1         # a
   b         2           5           2         # a,b
   c         5          15           2         # a,c (b has ended)
   d         8          13           3         # a,c,d
   e        12          20           3         # c,d,e (a has ended)
   f        21          30           1         # f (c,d,e have ended)

Answer 1

select      task_id,start_time,end_time,running_tasks 

from       (select      task_id,tm,op,start_time,end_time

                       ,sum(op) over 
                        (
                            order by    tm,op 
                            rows        unbounded preceding
                        ) as running_tasks 

            from       (select      task_id,start_time as tm,1 as op,start_time,end_time 
                        from        tasks 

                        union   all 

                        select      task_id,end_time as tm,-1 as op,start_time,end_time 
                        from        tasks 
                        ) t 
            )t 

where       op = 1
;

Answer 2

您可以使用相关子查询，在这种情况下是自联接；不需要分析函数。启用standard SQL（取消选中 UI 中“显示选项”下的“使用旧版 SQL”）后，您可以运行此示例：

WITH tasks AS (
  SELECT
    task_id,
    start_time,
    end_time
  FROM UNNEST(ARRAY<STRUCT<task_id STRING, start_time INT64, end_time INT64>>[
    ('a', 1, 10),
    ('b', 2, 5),
    ('c', 5, 15),
    ('d', 8, 13),
    ('e', 12, 20),
    ('f', 21, 30)
  ])
)
SELECT
  *,
  (SELECT COUNT(*) FROM tasks t2
   WHERE t.start_time >= t2.start_time AND
   t.start_time < t2.end_time) AS running_tasks
FROM tasks t
ORDER BY task_id;

Answer 3

正如 Elliott 所提到的 - “向新用户解释分析函数通常更困难”，即使是老用户也并不总是 100% 擅长它（虽然非常接近它）！
所以，虽然 Dudu Markovitz 的回答很棒——不幸的是，它仍然是不正确的（至少根据我对问题的理解）。不正确的情况是您在同一个 start_time 启动了多个任务 - 所以这些任务的“运行任务”结果错误

作为一个例子 - 考虑下面的例子：

task_id  start_time  end_time
   a         1          10
   aa        1           2
   aaa       1           8
   b         2           5
   c         5          15
   d         8          13
   e        12          20
   f        21          30

我想，你会期待以下结果：

task_id  start_time  end_time  running_tasks
   a         1          10           3         # a,aa,aaa
   aa        1           2           3         # a,aa,aaa
   aaa       1           8           3         # a,aa,aaa
   b         2           5           3         # a,aaa,b (aa has ended)
   c         5          15           3         # a,aaa,c (b has ended)
   d         8          13           3         # a,c,d (aaa has ended)
   e        12          20           3         # c,d,e (a has ended)
   f        21          30           1         # f (c,d,e have ended)     

如果你用 Dudu 的代码试试 - 你会得到下面的代替

task_id  start_time  end_time  running_tasks
   a         1          10           1        
   aa        1           2           2        
   aaa       1           8           3        
   b         2           5           3        
   c         5          15           3        
   d         8          13           3        
   e        12          20           3        
   f        21          30           1        

您可以看到任务 a 和 aa 的结果错误。
原因是因为使用了ROWS UNBOUNDED PRECEDING 而不是RANGE UNBOUNDED PRECEDING - 细微但非常重要的细微差别！

所以下面的查询会给你正确的结果

SELECT  task_id,start_time,end_time,running_tasks 
FROM  (
  SELECT  
    task_id, tm, op, start_time, end_time,
    SUM(op) OVER (ORDER BY  tm ,op RANGE UNBOUNDED PRECEDING) AS running_tasks 
  FROM  (
    SELECT  
      task_id, start_time AS tm, 1 AS op, start_time, end_time 
    FROM  tasks UNION  ALL 
    SELECT  
      task_id, end_time AS tm, -1 AS op, start_time, end_time 
    FROM  tasks 
  ) t 
)t 
WHERE  op = 1
ORDER BY start_time       

快速总结：
ROWS UNBOUNDED PRECEDING - 根据行的位置设置窗口框架
而
RANGE UNBOUNDED PRECEDING - 根据行值设置窗口框架

再次 - 正如 Elliott 所提到的 - 这比 JOIN 概念要复杂得多 - 但它值得（因为它比连接更有效） - 了解更多关于 Window Frame Clause 和 ROWS 与 RANGE 的使用