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有没有更好的方法来计算中位数(不是平均数)

aggregate-functions postgresql sql

Ghislain Leveque · 技术社区 · 14 年前

CREATE TABLE x (i serial primary key, value integer not null);

我想计算 value

下面是我如何计算中位数的方法,但我想一定有更好的方法:

SELECT AVG(values_around_median) AS median
  FROM (
    SELECT
       DISTINCT(CASE WHEN FIRST_VALUE(above) OVER w2 THEN MIN(value) OVER w3 ELSE MAX(value) OVER w2 END)
        AS values_around_median
      FROM (
        SELECT LAST_VALUE(value) OVER w AS value,
               SUM(COUNT(*)) OVER w > (SELECT count(*)/2 FROM x) AS above
          FROM x
          GROUP BY value
          WINDOW w AS (ORDER BY value)
          ORDER BY value
        ) AS find_if_values_are_above_or_below_median
      WINDOW w2 AS (PARTITION BY above ORDER BY value DESC),
             w3 AS (PARTITION BY above ORDER BY value ASC)
    ) AS find_values_around_median

7 回复 | 直到 13 年前

Scott Bailey 14 年前

确实有一个更简单的方法。在Postgres中,您可以定义自己的聚合函数。不久前,我在PostgreSQL代码段库中发布了一些函数来实现中间值、模式和范围。

http://wiki.postgresql.org/wiki/Aggregate_Median

Lukas Eder 10 年前

是的,在postgresql9.4中,您可以使用新引入的逆分布函数 PERCENTILE_CONT() ,一个在SQL标准中指定的有序集聚合函数。

WITH t(value) AS (
  SELECT 1   UNION ALL
  SELECT 2   UNION ALL
  SELECT 100 
)
SELECT
  percentile_cont(0.5) WITHIN GROUP (ORDER BY value)
FROM
  t;

This emulation of MEDIAN() via PERCENTILE_CONT() is also documented here

Erwin Brandstetter 10 年前

更简单的查询:

WITH y AS (
   SELECT value, row_number() OVER (ORDER BY value) AS rn
   FROM   x
   WHERE  value IS NOT NULL
   )
, c AS (SELECT count(*) AS ct FROM y) 
SELECT CASE WHEN c.ct%2 = 0 THEN
          round((SELECT avg(value) FROM y WHERE y.rn IN (c.ct/2, c.ct/2+1)), 3)
       ELSE
                (SELECT     value  FROM y WHERE y.rn = (c.ct+1)/2)
       END AS median
FROM   c;

要点

核心功能是 row_number() window function ,从版本8.4开始就有了
最后一个SELECT得到一行不均匀的数字和 avg()

测试表明,新版本比问题中的查询快4倍(并产生正确的结果):

CREATE TEMP TABLE x (value int);
INSERT INTO x SELECT generate_series(1,10000);
INSERT INTO x VALUES (NULL),(NULL),(NULL),(3);

Chris B 8 年前

对于谷歌用户来说:还有 http://pgxn.org/dist/quantile 中线可在安装此延长线后在一条线上计算。

Ghost 8 年前

仅具有本机postgres函数的简单sql:

select 
    case count(*)%2
        when 1 then (array_agg(num order by num))[count(*)/2+1]
        else ((array_agg(num order by num))[count(*)/2]::double precision + (array_agg(num order by num))[count(*)/2+1])/2
    end as median
from unnest(array[5,17,83,27,28]) num;

toha 8 年前

CREATE TABLE array_table (id integer, values integer[]) ;

INSERT INTO array_table VALUES ( 1,'{1,2,3}');
INSERT INTO array_table VALUES ( 2,'{4,5,6,7}');

select id, values, cardinality(values) as array_length,
(case when cardinality(values)%2=0 and cardinality(values)>1 then (values[(cardinality(values)/2)]+ values[((cardinality(values)/2)+1)])/2::float 
 else values[(cardinality(values)+1)/2]::float end) as median  
 from array_table

或者您可以创建一个函数,并在以后的查询中随时使用它。

CREATE OR REPLACE FUNCTION median (a integer[]) 
RETURNS float AS    $median$ 
Declare     
    abc float; 
BEGIN    
    SELECT (case when cardinality(a)%2=0 and cardinality(a)>1 then 
           (a[(cardinality(a)/2)] + a[((cardinality(a)/2)+1)])/2::float   
           else a[(cardinality(a)+1)/2]::float end) into abc;    
    RETURN abc; 
END;    
$median$ 
LANGUAGE plpgsql;

select id,values,median(values) from array_table

Sowmiya Raja Radhakrishnan 8 年前

使用下面的函数查找第n个百分位数

CREATE or REPLACE FUNCTION nth_percentil(anyarray, int)
    RETURNS 
        anyelement as 
    $$
        SELECT $1[$2/100.0 * array_upper($1,1) + 1] ;
    $$ 
LANGUAGE SQL IMMUTABLE STRICT;

你的情况是第50百分位。

SELECT nth_percentil(ARRAY (SELECT Field_name FROM table_name ORDER BY 1),50)

这将给你第50百分位,这是基本上的中位数。