BigQuery中重疊間隔的數量

Question

給定一個間隔表，我可以有效地查詢每個間隔開始時的當前打開間隔數（包括當前間隔本身）嗎？

例如，下表給出：

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

我想下面的輸出：

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

在小數據集，我可以對自己加入該數據集解決這個問題：

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 

對於大型數據集，CROSS JOIN既不切實際也不必要，因爲ny給出答案只取決於少數前面的區間（當按start_time排序時）。事實上，在我擁有的數據集中，它超時。有沒有更好的方法來實現這一目標？

Answer 1

... CROSS JOIN既不切實際也不必要...
有沒有更好的方法來實現這個目標？

請嘗試下面的BigQuery標準SQL。沒有加入參與

#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 

您可以測試/使用下面的例子發揮它從你的問題

#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