用MySQL计算中位数最简单(希望不会太慢)的方法是什么?我已经使用AVG(x)来寻找平均值，但我很难找到一个简单的方法来计算中位数。现在，我将所有的行返回到PHP，进行排序，然后选择中间的行，但是肯定有一些简单的方法可以在一个MySQL查询中完成它。
示例数据:
id | val
--------
1 4
2 7
3 2
4 2
5 9
6 8
7 3
对val排序得到2 2 3 4 7 8 9，因此中位数应该是4，而SELECT AVG(val) == 5。
当前回答
最简单和快速的方法来计算中位数在mysql。
select x.col
from (select lat_n,
count(1) over (partition by 'A') as total_rows,
row_number() over (order by col asc) as rank_Order
from station ft) x
where x.rank_Order = round(x.total_rows / 2.0, 0)
其他回答
我没有将这个解决方案的性能与这里发布的其他答案进行比较，但我发现这个解决方案是最容易理解的，并且涵盖了计算中位数的全部数学公式。换句话说，这个解决方案对于偶数和奇数数据集足够健壮:
SELECT CASE
-- odd-numbered data sets:
WHEN MOD(COUNT(*), 2) = 1 THEN (SELECT median.<value> AS median
FROM
(SELECT t1.<value>
FROM (SELECT <value>,
ROW_NUMBER() OVER(ORDER BY <value>) AS rownum
FROM <data>) t1,
(SELECT COUNT(*) AS num_records FROM <data>) t2
WHERE t1.rownum =(t2.num_records) / 2) as median)
-- even-numbered data sets:
ELSE (select (low_bound.<value> + up_bound.<value>) / 2 AS median
FROM
(SELECT t1.<value>
FROM (SELECT <value>,
ROW_NUMBER() OVER(ORDER BY <value>) AS rownum
FROM <data>) t1,
(SELECT COUNT(*) AS num_records FROM <data>) t2
WHERE t1.rownum =(t2.num_records - 1) / 2) as low_bound,
(SELECT t1.<value>
FROM (SELECT <value>,
ROW_NUMBER() OVER(ORDER BY <value>) AS rownum
FROM station) t1,
(SELECT COUNT(*) AS num_records FROM data) t2
WHERE t1.rownum =(t2.num_records + 1) / 2) as up_bound)
END
FROM <data>
我的代码，高效，没有表或额外的变量:
SELECT
((SUBSTRING_INDEX(SUBSTRING_INDEX(group_concat(val order by val), ',', floor(1+((count(val)-1) / 2))), ',', -1))
+
(SUBSTRING_INDEX(SUBSTRING_INDEX(group_concat(val order by val), ',', ceiling(1+((count(val)-1) / 2))), ',', -1)))/2
as median
FROM table;
不幸的是，无论是TheJacobTaylor还是velcrow的答案都不会返回当前版本MySQL的准确结果。
从上面来看，魔术贴的答案是接近的，但它不能正确计算具有偶数行数的结果集。中位数定义为1)奇数集上的中间数，或2)偶数集上两个中间数的平均值。
所以，这里是魔术贴的解决方案修补处理奇数和偶数集:
SELECT AVG(middle_values) AS 'median' FROM (
SELECT t1.median_column AS 'middle_values' FROM
(
SELECT @row:=@row+1 as `row`, x.median_column
FROM median_table AS x, (SELECT @row:=0) AS r
WHERE 1
-- put some where clause here
ORDER BY x.median_column
) AS t1,
(
SELECT COUNT(*) as 'count'
FROM median_table x
WHERE 1
-- put same where clause here
) AS t2
-- the following condition will return 1 record for odd number sets, or 2 records for even number sets.
WHERE t1.row >= t2.count/2 and t1.row <= ((t2.count/2) +1)) AS t3;
要使用它，请遵循以下3个简单步骤:
将上面代码中的“median_table”(出现2次)替换为您的表名 将“median_column”(3次)替换为您希望为其查找中位数的列名 如果你有一个WHERE条件，用WHERE条件替换“WHERE 1”(2次)
create table med(id integer);
insert into med(id) values(1);
insert into med(id) values(2);
insert into med(id) values(3);
insert into med(id) values(4);
insert into med(id) values(5);
insert into med(id) values(6);
select (MIN(count)+MAX(count))/2 from
(select case when (select count(*) from
med A where A.id<B.id)=(select count(*)/2 from med) OR
(select count(*) from med A where A.id>B.id)=(select count(*)/2
from med) then cast(B.id as float)end as count from med B) C;
?column?
----------
3.5
(1 row)
OR
select cast(avg(id) as float) from
(select t1.id from med t1 JOIN med t2 on t1.id!= t2.id
group by t1.id having ABS(SUM(SIGN(t1.id-t2.id)))=1) A;
你可以使用这里的用户定义函数。
