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

Question

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?

Answer 1

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.

Answer 2

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

Answer 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);

Answer 4

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

Answer 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.

Answer 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

Answer 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.