GROUP BY 和聚合连续数值

Question

使用 PostgreSQL 9.0。

假设我有一个包含字段的表：company、profession 和 year。我想返回一个包含独特公司和职业的结果，但根据数字序列聚合（到一个数组中即可）年份：

示例表：

+-----------------------------+
| company | profession | year |
+---------+------------+------+
| Google  | Programmer | 2000 |
| Google  | Sales      | 2000 |
| Google  | Sales      | 2001 |
| Google  | Sales      | 2002 |
| Google  | Sales      | 2004 |
| Mozilla | Sales      | 2002 |
+-----------------------------+

我对一个查询感兴趣，该查询将输出类似于以下内容的行：

+-----------------------------------------+
| company | profession | year             |
+---------+------------+------------------+
| Google  | Programmer | [2000]           |
| Google  | Sales      | [2000,2001,2002] |
| Google  | Sales      | [2004]           |
| Mozilla | Sales      | [2002]           |
+-----------------------------------------+

基本功能是只有连续年份才可以分组在一起。

Answer 1

@a_horse_with_no_name 的答案很有价值一个正确的解决方案，就像我在评论中已经说过的那样，作为学习如何在 PostgreSQL 中使用不同类型的窗口函数的好材料。

然而，我不禁感到，对于像这样的问题，该答案中所采取的方法有点太过了。基本上，在继续在数组中聚合年份之前，您需要的是一个附加的分组标准。您已经有了公司和职业，现在您只需要一些东西来区分属于不同序列的年份。

这正是上述答案所提供的，这正是我认为可以用更简单的方式完成的。方法如下：

WITH MarkedForGrouping AS (
  SELECT
    company,
    profession,
    year,
    year - ROW_NUMBER() OVER (
      PARTITION BY company, profession
      ORDER BY year
    ) AS seqID
  FROM atable
)
SELECT
  company,
  profession,
  array_agg(year) AS years
FROM MarkedForGrouping
GROUP BY
  company,
  profession,
  seqID

Answer 2

使用 PL/pgSQL 的程序解决方案

对于具有聚合/Windows 函数的普通 SQL 来说，这个问题相当难以处理。虽然循环通常比使用纯 SQL 的基于集合的解决方案慢，但使用 PL/pgSQL 的过程解决方案可以通过对表进行单次顺序扫描（FOR 循环），并且在这种特殊情况下应该明显更快：

测试表：

CREATE TEMP TABLE tbl (company text, profession text, year int);
INSERT INTO tbl VALUES
  ('Google',  'Programmer', 2000)
, ('Google',  'Sales',      2000)
, ('Google',  'Sales',      2001)
, ('Google',  'Sales',      2002)
, ('Google',  'Sales',      2004)
, ('Mozilla', 'Sales',      2002)
;

函数：

CREATE OR REPLACE FUNCTION f_periods()
  RETURNS TABLE (company text, profession text, years int[])
  LANGUAGE plpgsql AS
$func$
DECLARE
   r  tbl; -- use table type as row variable
   r0 tbl;
BEGIN
   FOR r IN
      SELECT * FROM tbl t ORDER BY t.company, t.profession, t.year
   LOOP
      IF ( r.company,  r.profession,  r.year)
      <> (r0.company, r0.profession, r0.year + 1) THEN -- not true for first row

         RETURN QUERY
         SELECT r0.company, r0.profession, years; -- output row

         years := ARRAY[r.year];     -- start new array
      ELSE
         years := years || r.year;   -- add to array - year can be NULL, too
      END IF;

      r0 := r;                       -- remember last row
   END LOOP;

   RETURN QUERY                      -- output last iteration
   SELECT r0.company, r0.profession, years;
END
$func$;

调用：

SELECT * FROM f_periods();

db>>fiddle 此处

生成请求的结果。

Answer 3

识别非连续值总是有点棘手，并且涉及多个嵌套子查询（至少我无法想出更好的解决方案）。

第一步是识别年份的非连续值：

步骤 1) 识别非连续值

select company, 
       profession,
       year,
       case 
          when row_number() over (partition by company, profession order by year) = 1 or 
               year - lag(year,1,year) over (partition by company, profession order by year) > 1 then 1
          else 0
       end as group_cnt
from qualification

这将返回以下结果：

 company | profession | year | group_cnt
---------+------------+------+-----------
 Google  | Programmer | 2000 |         1
 Google  | Sales      | 2000 |         1
 Google  | Sales      | 2001 |         0
 Google  | Sales      | 2002 |         0
 Google  | Sales      | 2004 |         1
 Mozilla | Sales      | 2002 |         1

现在，使用 group_cnt 值，我们可以为具有连续年份的每个组创建“组 ID”：

步骤 2 ) 定义组 ID

select company,
   profession,
   year,
   sum(group_cnt) over (order by company, profession, year) as group_nr
from ( 
select company, 
       profession,
       year,
       case 
          when row_number() over (partition by company, profession order by year) = 1 or 
               year - lag(year,1,year) over (partition by company, profession order by year) > 1 then 1
          else 0
       end as group_cnt
from qualification
) t1

这将返回以下结果：

 company | profession | year | group_nr
---------+------------+------+----------
 Google  | Programmer | 2000 |        1
 Google  | Sales      | 2000 |        2
 Google  | Sales      | 2001 |        2
 Google  | Sales      | 2002 |        2
 Google  | Sales      | 2004 |        3
 Mozilla | Sales      | 2002 |        4
(6 rows)

如您所见，每个“组”都有自己的 group_nr，我们最终可以通过添加另一个派生表来使用它进行聚合：

步骤 3) 最终查询

select company,
       profession,
       array_agg(year) as years
from (
  select company,
       profession,
       year,
       sum(group_cnt) over (order by company, profession, year) as group_nr
  from ( 
    select company, 
           profession,
           year,
           case 
              when row_number() over (partition by company, profession order by year) = 1 or 
                   year - lag(year,1,year) over (partition by company, profession order by year) > 1 then 1
              else 0
           end as group_cnt
    from qualification
  ) t1
) t2
group by company, profession, group_nr
order by company, profession, group_nr

以下结果：

 company | profession |      years
---------+------------+------------------
 Google  | Programmer | {2000}
 Google  | Sales      | {2000,2001,2002}
 Google  | Sales      | {2004}
 Mozilla | Sales      | {2002}
(4 rows)

这将返回 正是如果我没记错的话，你想要什么。