BigQuery 计算两个日期范围重叠

Question

我最终通过下载所有数据并在 Python 中遍历它来解决了这个问题，但我想知道是否有办法在 BigQuery 中做到这一点。

我们有一个包含开始日期和结束日期的表格：

begin_date, end_date
'2016-02-19', '2016-02-19'
'2016-02-20', '2016-02-25'
'2016-02-21', '2016-02-25'
'2016-02-22', NULL

我们想要 begin_date

SELECT COUNT(*) FROM `table` WHERE begin_date <= '2016-12-19' AND (end_date >= '2016-12-19' OR end_date IS NULL)

因此，如果我为每个感兴趣的值手动执行此操作，所需的输出可能如下所示：

begin_date, count
2016-02-19, 1
2016-02-20, 1
2016-02-21, 2
2016-02-22, 3
2016-02-23, 3
2016-02-24, 3
2016-02-25, 3
2016-02-26, 1
etc.

创建要迭代的日期列表很容易：

WITH dates AS (SELECT * FROM UNNEST(GENERATE_DATE_ARRAY('2018-10-01', '2020-09-30', INTERVAL 1 DAY)) AS example)

现在我正在努力在所有这些日期中应用上述 WHERE 子句。我看到在匹配单个列 (like here) 时，具有范围的分区是如何工作的，但我需要同时匹配 begin_date 和 end_date。

我认为我可以这样做：

SELECT
  status_begin_date,
  (SELECT COUNT(1) FROM UNNEST(ends) AS e WHERE (e >= status_begin_date OR e IS NULL)) AS cnt
FROM (
  SELECT
    status_begin_date,
    ARRAY_AGG(status_end_date) OVER(ORDER BY status_begin_date) AS ends
  FROM `table`
)
ORDER BY status_begin_date

取自here。这适用于 StackOverflow 答案中给出的小示例，但我在有几亿行的表上使用它时遇到资源错误： BigQuery 中是否有可扩展的解决方案？

Answer 1

以下是 BigQuery 标准 SQL，不使用低效游标方法，而是使用基于经典 sql 集的方法

#standardSQL
WITH dates AS (
  SELECT day 
  FROM (SELECT MIN(begin_date) min_date, MAX(end_date) max_date FROM `table`), 
  UNNEST(GENERATE_DATE_ARRAY(min_date, CURRENT_DATE(), INTERVAL 1 DAY)) AS day
)
SELECT day, COUNT(*) 
FROM dates 
JOIN `table` 
ON begin_date <= day AND (end_date >= day OR end_date IS NULL)
GROUP BY day

您可以使用您问题中的示例数据进行测试，如以下示例所示

#standardSQL
WITH `table` AS (
  SELECT DATE '2016-02-19' begin_date, DATE '2016-02-19' end_date UNION ALL
  SELECT '2016-02-20', '2016-02-25' UNION ALL
  SELECT '2016-02-21', '2016-02-25' UNION ALL
  SELECT '2016-02-22', NULL
), dates AS (
  SELECT day 
  FROM (SELECT MIN(begin_date) min_date, MAX(end_date) max_date FROM `table`), 
  UNNEST(GENERATE_DATE_ARRAY(min_date, max_date, INTERVAL 1 DAY)) AS day
)
SELECT day, COUNT(*) 
FROM dates 
JOIN `table` 
ON begin_date <= day AND (end_date >= day OR end_date IS NULL)
GROUP BY day
-- ORDER BY day  

结果

Row day         f0_  
1   2016-02-19  1    
2   2016-02-20  1    
3   2016-02-21  2    
4   2016-02-22  3    
5   2016-02-23  3    
6   2016-02-24  3    
7   2016-02-25  3    

Answer 2

这个讨厌的代码起作用了：

DECLARE dates ARRAY <DATE>;
DECLARE x INT64 DEFAULT 0;
DECLARE results ARRAY <INT64>;
DECLARE results_dates ARRAY <DATE>;
DECLARE result INT64;
DECLARE date DATE;
SET dates = GENERATE_DATE_ARRAY('2016-02-17', '2019-05-13', INTERVAL 1 DAY);
LOOP
  SET date = dates[OFFSET(x)];
  SET result = (SELECT COUNT(*) FROM `table` WHERE begin_date <= date AND (end_date >= date OR end_date IS NULL));
  SET results = ARRAY_CONCAT(results, [result]);
  SET results_dates = ARRAY_CONCAT(results_dates, [date]);
  SET x = x + 1;
  IF x >= ARRAY_LENGTH(dates) THEN
    LEAVE;
  END IF;
END LOOP;
SELECT date, count_subscribers
FROM UNNEST(results_dates) AS date WITH OFFSET 
JOIN UNNEST(results) AS count_subscribers WITH OFFSET
USING(OFFSET)

运行时间为 1.5 小时，比我的 Python 代码（7 小时）要好，但 BigQuery 代码不可并行化，而 Python 代码可以。