1

我在 BigQuery 中有一个具有以下结构的表：

datetime | event  | value
==========================
1        | add    | 1
---------+--------+-------
2        | remove | 1
---------+--------+-------
6        | add    | 2
---------+--------+-------
8        | add    | 3
---------+--------+-------
11       | add    | 4
---------+--------+-------
23       | remove | 3
---------+--------+-------

我正在尝试构建一个视图，该视图list向包含数组当前状态的每一行添加一列。数组永远不会包含重复项。这应该是结果：

datetime | event  | value | list
===================================
1        | add    | 1     | [1]
---------+--------+-------+--------
2        | remove | 1     | []
---------+--------+-------+--------
6        | add    | 2     | [2]
---------+--------+-------+--------
8        | add    | 3     | [2,3]
---------+--------+-------+--------
11       | add    | 4     | [2,3,4]
---------+--------+-------+--------
23       | remove | 3     | [2,4]
---------+--------+-------+--------

我尝试使用分析函数，但没有成功。使用数组的 api 非常有限。如果我可以使用递归WITH子句，我想我会成功，不幸的是，这在 BigQuery 中是不可能的。

我正在使用启用了标准 SQL 的 BigQuery。

4

4 回答 4

3

以下是 BigQuery SQL（可能只是众多选项之一） 

#standardSQL
CREATE TEMP FUNCTION CUST_ARRAY_AGG(arr ARRAY<STRUCT<event STRING, value STRING>>)
RETURNS ARRAY<STRING>
LANGUAGE js AS """
  var result = [];  
  for (i = 0; i < arr.length; i++) { 
    if (arr[i].event == 'add') {
      result.push(arr[i].value);
    } else {
      var index = result.indexOf(arr[i].value);
      if (index > -1) {
        result.splice(index, 1);
      }
    }
  }
  return result;
""";
WITH `project.dataset.events` AS (
  SELECT 1 dt, 'add' event, '1' value UNION ALL
  SELECT 2, 'remove', '1' UNION ALL
  SELECT 6, 'add', '2' UNION ALL
  SELECT 8, 'add', '3' UNION ALL
  SELECT 11, 'add', '4' UNION ALL
  SELECT 23, 'remove', '3' 
)
SELECT dt, event, value, 
  CUST_ARRAY_AGG(arr) list_as_arr,
  (SELECT CONCAT('[', IFNULL(STRING_AGG(item), ''), ']') FROM UNNEST(CUST_ARRAY_AGG(arr)) item) list_as_string
FROM (
  SELECT dt, event, value,
    ARRAY_AGG(STRUCT<event STRING, value STRING>(event, value)) OVER(ORDER BY dt) arr
  FROM `project.dataset.events`
)

结果如下 

Row dt  event   value   list_as_arr list_as_string   
1   1   add     1       1           [1]  
2   2   remove  1                   []   
3   6   add     2       2           [2]  
4   8   add     3       2           [2,3]    
                        3        
5   11  add     4       2           [2,3,4]  
                        3        
                        4        
6   23  remove  3       2           [2,4]    
                        4
于 2018-02-14T21:13:23.300 回答
3

以下版本适用于 BigQuery 标准 SQL，仅使用纯 SQL（无 JS UDF） 

#standardSQL
WITH `project.dataset.events` AS (
  SELECT 1 dt,'add' event,'1' value UNION ALL
  SELECT 2,   'remove',   '1' UNION ALL
  SELECT 6,   'add',      '2' UNION ALL
  SELECT 8,   'add',      '3' UNION ALL
  SELECT 11,  'add',      '4' UNION ALL
  SELECT 23,  'remove',   '3' 
), cum AS (
  SELECT dt, event, value,
    SUM(IF(event = 'add', 1, -1)) OVER(PARTITION BY value ORDER BY dt) state
  FROM `project.dataset.events`
), pre AS (
  SELECT 
    a.dt, a.event, a.value, a.state, b.value AS b_value,
    ARRAY_AGG(b.state ORDER BY b.dt DESC)[SAFE_OFFSET(0)] b_state, 
    MAX(b.dt) b_dt 
  FROM cum a
  JOIN cum b ON b.dt <= a.dt
  GROUP BY a.dt, a.event, a.value, a.state, b.value
)
SELECT dt, event, value, 
  SPLIT(IFNULL(STRING_AGG(IF(b_state = 1, b_value, NULL) ORDER BY b_dt), '')) list_as_array,
  CONCAT('[', IFNULL(STRING_AGG(IF(b_state = 1, b_value, NULL) ORDER BY b_dt), ''), ']') list_as_string
FROM pre
GROUP BY dt, event, value
ORDER BY dt

结果“令人惊讶”：o）与我之前回答/发布的 JS UDF 版本完全相同 

Row dt  event   value   list_as_arr list_as_string   
1   1   add     1       1           [1]  
2   2   remove  1                   []   
3   6   add     2       2           [2]  
4   8   add     3       2           [2,3]    
                        3        
5   11  add     4       2           [2,3,4]  
                        3        
                        4        
6   23  remove  3       2           [2,4]    
                        4

注意：我认为上面的设计可能有点过度 - 但我只是没有时间可能完善/优化它 - 应该是可行的 - 把这个留给问题的所有者

于 2018-02-15T01:28:21.737 回答
1

虽然已经解决了我喜欢这个问题。此解决方案中的想法是首先使用窗口获取完整历史记录，同时考虑所有前面的行 - 然后删除删除项：

#standardSQL
-- stolen test table ;)
WITH test AS (
  SELECT 1 dt,'add' event,'1' value UNION ALL
  SELECT 2,   'remove',   '1' UNION ALL
  SELECT 6,   'add',      '2' UNION ALL
  SELECT 8,   'add',      '3' UNION ALL
  SELECT 11,  'add',      '4' UNION ALL
  SELECT 23,  'remove',   '3' 
)

, windowing as (
SELECT *,
  -- add history using window function
  ARRAY_AGG(STRUCT(event, value)) OVER ( ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) as history
FROM test)

SELECT 
  dt,
  event,
  value,
  --history, -- for testing

  -- Get all added items that are not removed items
  -- This sub-select runs within the row only, treating history-array as (sub-)table
  (SELECT ARRAY_AGG(value) value FROM unnest(t.history) l 
    WHERE l.event = 'add' 
    AND l.value NOT IN 
      (SELECT l.value FROM unnest(t.history) l WHERE l.event = 'remove')
  ) AS list
FROM windowing t

我没有比较性能，但会感兴趣！

于 2018-02-15T11:49:47.880 回答
0

所有原始答案都很棒（尤其是我的 - 大声笑）并且主要基于基于集合（处理大数据的最佳方式）处理，在这种情况下对于非 sql 用户来说可能变得足够复杂！

幸运的是，对脚本和存储过程的支持现在处于测试阶段（截至 2019 年 10 月）

您可以提交多个用分号分隔的语句，BigQuery 现在可以运行它们。并且以程序方式而不是基于集合的方式表达所需的逻辑（显然至少以性能为代价）要容易得多-因此更多用户可以受益

下面的脚本实现了问题中表达的逻辑 

DECLARE arr ARRAY<STRUCT<dt INT64, event STRING, value STRING>>;
DECLARE result ARRAY<STRUCT<dt INT64, event STRING, value STRING, list STRING>> DEFAULT [STRUCT(NULL, '', '', '')];
DECLARE list ARRAY<STRING> DEFAULT [];
DECLARE i, m INT64 DEFAULT -1; 

SET arr = (
  WITH t AS (
    SELECT 1 dt,'add' event,'1' value UNION ALL
    SELECT 2,   'remove',   '1' UNION ALL
    SELECT 6,   'add',      '2' UNION ALL
    SELECT 8,   'add',      '3' UNION ALL
    SELECT 11,  'add',      '4' UNION ALL
    SELECT 23,  'remove',   '3' 
  )
  SELECT ARRAY_AGG(t) FROM t
);

SET m = ARRAY_LENGTH(arr);

LOOP
  SET i = i + 1;
  IF i >= m THEN LEAVE;
  ELSE
    IF arr[OFFSET(i)].event = 'add' THEN 
      SET list = (
        SELECT ARRAY_CONCAT(list, [arr[OFFSET(i)].value])
      );    
    ELSE
      SET list = ARRAY(
        SELECT item
        FROM UNNEST(list) item 
        WHERE item != arr[OFFSET(i)].value
      );    
    END IF;

    SET result = (
      SELECT ARRAY_CONCAT(
        result, 
        [(arr[OFFSET(i)].dt, arr[OFFSET(i)].event, arr[OFFSET(i)].value, ARRAY_TO_STRING(list, ','))]
      )
    );
  END IF;
END LOOP;

SELECT * FROM UNNEST(result) WHERE NOT dt IS NULL;

结果

在此处输入图像描述

于 2019-10-04T19:57:30.880 回答