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
安装和使用本mysql统计函数:http://www.xarg.org/2012/07/statistical-functions-in-mysql/
之后，计算中值就很简单了:
SELECT median(val) FROM data;
其他回答
这是我的办法。当然，你可以把它放到一个过程中:-)
SET @median_counter = (SELECT FLOOR(COUNT(*)/2) - 1 AS `median_counter` FROM `data`);
SET @median = CONCAT('SELECT `val` FROM `data` ORDER BY `val` LIMIT ', @median_counter, ', 1');
PREPARE median FROM @median;
EXECUTE median;
你可以避免变量@median_counter，如果你替换它:
SET @median = CONCAT( 'SELECT `val` FROM `data` ORDER BY `val` LIMIT ',
(SELECT FLOOR(COUNT(*)/2) - 1 AS `median_counter` FROM `data`),
', 1'
);
PREPARE median FROM @median;
EXECUTE median;
我使用下表的解决方案在MySQL:
CREATE TABLE transactions (
transaction_id int , user_id int , merchant_name varchar(255), transaction_date date , amount int
);
INSERT INTO transactions (transaction_id, user_id, merchant_name, transaction_date, amount)
VALUES (1, 1 ,'abc', '2015-08-17', 100),(2, 2, 'ced', '2015-2-17', 100),(3, 1, 'def', '2015-2-16', 121),
(4, 1 ,'ced', '2015-3-17', 110),(5, 1, 'ced', '2015-3-17', 150),(6, 2 ,'abc', '2015-4-17', 130),
(7, 3 ,'ced', '2015-12-17', 10),(8, 3 ,'abc', '2015-8-17', 100),(9, 2 ,'abc', '2015-12-17', 140),(10, 1,'abc', '2015-9-17', 100),
(11, 1 ,'abc', '2015-08-17', 121),(12, 2 ,'ced', '2015-12-23', 130),(13, 1 ,'def', '2015-12-23', 13),(3, 4, 'abc', '2015-2-16', 120),(3, 4, 'def', '2015-2-16', 121),(3, 4, 'ced', '2015-2-16', 121);
计算“金额”列的中位数:
WITH Numbered AS
(
SELECT *, COUNT(*) OVER () AS TotatRecords,
ROW_NUMBER() OVER (ORDER BY amount) AS RowNum
FROM transactions
)
SELECT Avg(amount)
FROM Numbered
WHERE RowNum IN ( FLOOR((TotatRecords+1)/2), FLOOR((TotatRecords+2)/2) )
;
TotalRecords = 16 and Median = 120.5000
此查询将适用于两种情况，即偶数和奇数记录。
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;
我建议一个更快的方法。
获取行数:
SELECT CEIL(COUNT(*)/2);
然后取排序子查询的中间值:
SELECT max(val) FROM (SELECT val FROM data ORDER BY val limit @middlevalue);
我用5x10e6的随机数数据集进行了测试，它将在10秒内找到中位数。
下面的SQL代码将帮助您使用用户定义的变量来计算MySQL中的中位数。
create table employees(salary int); insert into employees values(8); insert into employees values(23); insert into employees values(45); insert into employees values(123); insert into employees values(93); insert into employees values(2342); insert into employees values(2238); select * from employees; Select salary from employees order by salary; set @rowid=0; set @cnt=(select count(*) from employees); set @middle_no=ceil(@cnt/2); set @odd_even=null; select AVG(salary) from (select salary,@rowid:=@rowid+1 as rid, (CASE WHEN(mod(@cnt,2)=0) THEN @odd_even:=1 ELSE @odd_even:=0 END) as odd_even_status from employees order by salary) as tbl where tbl.rid=@middle_no or tbl.rid=(@middle_no+@odd_even);
如果你正在寻找详细的解释，请参考这个博客。
