查询 table 与 array_agg/median 的所有先前位置，LAST_10，LAST_50，不包括当前位置

Question

这是我发布的这个回答得很好的问题的变体 :

我有一个数据库 table 有：

id | date       | position | name
--------------------------------------
1  | 2016-06-29 | 9        | Ben Smith
2  | 2016-06-29 | 1        | Ben Smith
3  | 2016-06-29 | 5        | Ben Smith
4  | 2016-06-29 | 6        | Ben Smith
5  | 2016-06-30 | 2        | Ben Smith
6  | 2016-06-30 | 2        | Tom Brown
7  | 2016-06-29 | 4        | Tom Brown
8  | 2016-06-30 | 2        | Tom Brown
9  | 2016-06-30 | 1        | Tom Brown

如何有效地查询 table 以便我可以使用 array_agg() 获取新列。

我已经尝试过以下查询，但是它非常慢而且错误，因为它没有按名称列对 previous_positions 进行分组：

 SELECT runners.id AS runner_id,
    btrim(regexp_replace(replace(upper(runners.name::text), '.'::text, ''::text), '[[:digit:]]'::text, ''::text, 'g'::text)) AS name,
    runners.position_two,
    (array_agg(runners.position_two) OVER w AS results
   FROM runners
  WINDOW w AS (PARTITION BY (btrim(regexp_replace(replace(upper(runners.name::text), '.'::text, ''::text), '[[:digit:]]'::text, ''::text, 'g'::text))) ORDER BY runners.id ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING);

我希望 table 输出看起来像这样

id | date       | position | name      | previous   | med  |med_20
----------------------------------------------------------------------
1  | 2016-06-29 | 9        | Ben Smith | {}         |      |
2  | 2016-06-29 | 1        | Ben Smith | {9}        | 9    | 9
3  | 2016-06-29 | 5        | Ben Smith | {9,1}      | 5    | 5
4  | 2016-06-29 | 6        | Ben Smith | {9,1,5}    | 5    | 5
5  | 2016-06-30 | 2        | Ben Smith | {9,1,5,6}  | 5.5  | 5.5
6  | 2016-06-30 | 2        | Tom Brown | {}         | None | None
7  | 2016-06-29 | 4        | Tom Brown | {2}        | 2    | 2
8  | 2016-06-30 | 2        | Tom Brown | {2,4}      | 3    | 3
9  | 2016-06-30 | 1        | Tom Brown | {2,4,2}    | 2    | 2

Answer 1

Postgres 没有 MEDIAN 的 built-in 聚合函数。但是，您可以使用 Postgres wiki. This snippet is also part of the ulib_agg user-defined library.

中提供的函数片段创建一个

创建后，您可以像 SUM 或 STRING_AGG 等具有类似 window 规范的任何聚合函数一样使用它。 Postgres 为您提供了为以逗号分隔的聚合函数指定多个 window 定义的选项。

因此，要获取前 20 条记录的 MEDIAN，您的 window 可以定义为在此查询中。

SELECT 
j.* ,  array_agg(position) over w as previous_positions,
       median(position)    over w_20 as med_20
  FROM jockeys j
WINDOW w as
(  partition by name ORDER BY id rows between 
     unbounded preceding and 1 preceding
     ),
     w_20 as
     (  partition by name ORDER BY id rows between 
           20 preceding and 1 preceding
     )

此外，如果您想截断小数位，您可以应用 ROUND 函数。

DEMO

查询 table 与 array_agg/median 的所有先前位置，LAST_10，LAST_50，不包括当前位置

Query table with array_agg/median of ALL previous positions, LAST_10, LAST_50, excluding current position

postgresql

array-agg