Question

这是我发布的previously这个精妙答案的变体：

我有一个数据库表，其中：

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

如何有效查询表，以便可以使用array_agg（）获取新列。

我已经尝试了以下查询，但是它的运行速度非常慢，而且也出错，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);

我希望表格输出看起来像这样

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的内置聚合函数。但是，您可以使用Postgres wiki中提供的功能代码段创建一个。此摘录也是ulib_agg user-defined library的一部分。

创建后，您可以像使用类似SUM规范的STRING_AGG或window之类的聚合函数一样使用它。 Postgres为您提供了为逗号分隔的聚合函数指定多个window定义的选项。

因此，要获得前20条记录的MEDIAN，可以按照此查询的定义窗口。

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，如果要截断十进制数字，可以应用ROUND函数。

DEMO

查询表，其中array_agg /中位数为所有先前位置，LAST_10，LAST_50，当前位置除外

1 个答案: