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
我有下面的代码，我在HackerRank上找到的，它很简单，适用于每一种情况。
SELECT M.MEDIAN_COL FROM MEDIAN_TABLE M WHERE
(SELECT COUNT(MEDIAN_COL) FROM MEDIAN_TABLE WHERE MEDIAN_COL < M.MEDIAN_COL ) =
(SELECT COUNT(MEDIAN_COL) FROM MEDIAN_TABLE WHERE MEDIAN_COL > M.MEDIAN_COL );
其他回答
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)
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。
set @r = 0;
select
case when mod(c,2)=0 then round(sum(lat_N),4)
else round(sum(lat_N)/2,4)
end as Med
from
(select lat_N, @r := @r+1, @r as id from station order by lat_N) A
cross join
(select (count(1)+1)/2 as c from station) B
where id >= floor(c) and id <=ceil(c)
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)
一个简单的方法来计算中位数在MySQL
set @ct := (select count(1) from station);
set @row := 0;
select avg(a.val) as median from
(select * from table order by val) a
where (select @row := @row + 1)
between @ct/2.0 and @ct/2.0 +1;
