Question

就像这里的示例一样，我想跨BigQuery数组进行计数：Distinct Count across Bigquery arrays

但是，我还有一些其他要求，这些要求使该帖子中提供的解决方案对我而言是可行的：

解决方案必须不使用UDF（太慢）
解决方案必须不使用HLL功能（必须准确）
解决方案必须不使用linked解决方案上显示的SELECT模式中的SELECT，因为该解决方案需要汇总在最终选择的一组灵活尺寸上用户使用BI工具

因此，尽管此扩展示例（包含用户作为分组维度）可以使用HLL：

#standardSQL
WITH
  test AS (
  SELECT
    'A' AS User, DATE('2018-01-01') AS ReportDate, 2 AS value, [1,2,3] AS key
  UNION ALL
  SELECT
    'A' AS User, DATE('2018-01-02') AS ReportDate, 3 AS value, [1,4,5] AS key
  UNION ALL
  SELECT
    'B' AS User, DATE('2018-01-02') AS ReportDate, 4 AS value, [4,5,6,7,8] AS key
  UNION ALL
  SELECT
    'B' AS User, DATE('2018-01-02') AS ReportDate, 5 AS value, [3,4,5,6,7] AS key )
SELECT
  User,
  SUM(value) total_value,
  HLL_COUNT.MERGE((
    SELECT
      HLL_COUNT.INIT(key)
    FROM
      UNNEST(key) key)) AS unique_key_count
FROM
  test
GROUP BY
  user

我需要一个能够满足上述要求的不同聚合数组计数版本。

同样，这意味着如果我仅将 ReportDate 分组，将 User / ReportDate 组合在一起，或者将该示例扩展了其他维度的情况下，它也应该可以正常工作

Answer 1

#standardSQL
WITH test AS
(
  SELECT 'A' AS User, DATE('2018-01-01') AS ReportDate, 2 AS value, [1,2,3] AS key UNION ALL
  SELECT 'A' AS User, DATE('2018-01-02') AS ReportDate, 3 AS value, [1,4,5] AS key UNION ALL
  SELECT 'B' AS User, DATE('2018-01-02') AS ReportDate, 4 AS value, [4,5,6,7,8] AS key UNION ALL
  SELECT 'B' AS User, DATE('2018-01-02') AS ReportDate, 5 AS value, [3,4,5,6,7] AS key  
)
SELECT 
  User,
  SUM(IF(flag=0, value, 0)) total_value,
  COUNT(DISTINCT key) unique_key_count
FROM test, UNNEST(key) key WITH OFFSET flag
GROUP BY User

有结果

Row User    total_value unique_key_count     
1   A       5           5    
2   B       9           6

在不使用HLL或UDF的情况下跨Bigquery数组进行简单区分计数

1 个答案: