基于使用BigQuery的给定条件,column_2中每个不同值的column_1值的计数

问题描述 投票:0回答:1

我有一张桌子,

date_1 | date_2
2018-12-08 | 2018-12-07
2018-12-09 | 2018-12-07
2018-12-13 | 2018-12-07
2018-12-16 | 2018-12-07
2018-12-14 | 2018-12-09

输出表应该是,

date_2 | count_of_date_1_after_date_2
2018-12-07 | 5
2018-12-09 | 3

我想在'date_1之后'计算date_2'。

我尝试了并得出了这个答案,但我不确定这是否是最好的方法。有一个更好的方法吗?

WITH

sample_table AS (
  SELECT DATE('2018-12-08') AS date_1, DATE('2018-12-07') AS date_2, 'AAA' as uid UNION ALL
  SELECT DATE('2018-12-09') AS date_1, DATE('2018-12-07') AS date_2, 'AAA' as uid  UNION ALL
  SELECT DATE('2018-12-13') AS date_1, DATE('2018-12-07') AS date_2, 'AAA' as uid  UNION ALL
  SELECT DATE('2018-12-16') AS date_1, DATE('2018-12-07') AS date_2, 'AAA' as uid  UNION ALL
  SELECT DATE('2018-12-14') AS date_1, DATE('2018-12-09') AS date_2, 'AAA' as uid 
),

distinct_date_2 AS (
  SELECT DISTINCT(date_2) AS distinct_date, uid FROM sample_table
)


SELECT distinct_date, COUNTIF(date_1 > distinct_date)
FROM sample_table
LEFT JOIN distinct_date_2
USING (uid)
GROUP BY distinct_date
ORDER BY distinct_date
google-bigquery standard-sql
1个回答
2
投票

以下是BigQuery Standard SQL

#standardSQL
SELECT date_2,
  (SELECT COUNT(1) FROM t.arr date_1 WHERE date_1 > date_2) AS count_of_date_1_after_date_2
FROM (
  SELECT date_2, ANY_VALUE(arr) arr
  FROM (
    SELECT date_2, ARRAY_AGG(date_1) OVER() arr
    FROM `project.dataset.table`
  ) 
  GROUP BY date_2
) t

您可以使用问题中的示例数据进行测试,使用上面的示例,如下例所示

#standardSQL
WITH `project.dataset.table` AS (
  SELECT DATE '2018-12-08' date_1, DATE '2018-12-07' date_2 UNION ALL
  SELECT '2018-12-09', '2018-12-07' UNION ALL
  SELECT '2018-12-13', '2018-12-07' UNION ALL
  SELECT '2018-12-16', '2018-12-07' UNION ALL
  SELECT '2018-12-14', '2018-12-09' 
)
SELECT date_2,
  (SELECT COUNT(1) FROM t.arr date_1 WHERE date_1 > date_2) AS count_of_date_1_after_date_2
FROM (
  SELECT date_2, ANY_VALUE(arr) arr
  FROM (
    SELECT date_2, ARRAY_AGG(date_1) OVER() arr
    FROM `project.dataset.table`
  ) 
  GROUP BY date_2
) t

结果

Row     date_2          count_of_date_1_after_date_2     
1       2018-12-07      5    
2       2018-12-09      3    
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