Is there a better way to calculate the median (not average)

问题

Suppose I have the following table definition:

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

I want to calculate the MEDIAN of value (not the AVG). The median is a value that divides the set in two subsets containing the same number of elements. If the number of elements is even, the median is the average of the biggest value in the lowest segment and the lowest value of the biggest segment. (See wikipedia for more details.)

Here is how I manage to calculate the MEDIAN but I guess there must be a better way:

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

Any ideas?

回答1:

Indeed there IS an easier way. In Postgres you can define your own aggregate functions. I posted functions to do median as well as mode and range to the PostgreSQL snippets library a while back.

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

回答2:

Yes, with PostgreSQL 9.4, you can use the newly introduced inverse distribution function PERCENTILE_CONT(), an ordered-set aggregate function that is specified in the SQL standard as well.

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.

回答3:

A simpler query for that:

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;

Major points

Ignores NULL values.
Core feature is the row_number() window function, which has been there since version 8.4
The final SELECT gets one row for uneven numbers and avg() of two rows for even numbers. Result is numeric, rounded to 3 decimal places.

Test shows, that the new version is 4x faster than (and yields correct results, unlike) the query in the question:

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

回答4:

For googlers: there is also http://pgxn.org/dist/quantile Median can be calculated in one line after installation of this extension.

回答5:

Simple sql with native postgres functions only:

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;

Sure you can add coalesce() or something if you want to handle nulls.

回答6:

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

Or you can create a function and use it any where in your further queries.

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

回答7:

Use the Below function for Finding nth percentile

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;

In Your case it's 50th Percentile.

Use the Below Query to get the Median

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

This will give you 50th percentile which is the median basically.

Hope this is helpful.

来源：https://stackoverflow.com/questions/3735252/is-there-a-better-way-to-calculate-the-median-not-average

标签

sql

postgresql

aggregate-functions