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
您可以使用窗口函数row_number()来回答查询以找到介质
select val
from (select val, row_number() over (order by val) as rownumber, x.cnt
from data, (select count(*) as cnt from data) x) abc
where rownumber=ceil(cnt/2);
其他回答
另一个对Velcrow答案的重复，但使用了一个中间表，并利用了用于行编号的变量来获得计数，而不是执行额外的查询来计算它。还开始计数，以便第一行是第0行，以便简单地使用Floor和Ceil选择中位数行。
SELECT Avg(tmp.val) as median_val
FROM (SELECT inTab.val, @rows := @rows + 1 as rowNum
FROM data as inTab, (SELECT @rows := -1) as init
-- Replace with better where clause or delete
WHERE 2 > 1
ORDER BY inTab.val) as tmp
WHERE tmp.rowNum in (Floor(@rows / 2), Ceil(@rows / 2));
关心奇数的计数-给出中间两个值的平均值。
SELECT AVG(val) FROM
( SELECT x.id, x.val from data x, data y
GROUP BY x.id, x.val
HAVING SUM(SIGN(1-SIGN(IF(y.val-x.val=0 AND x.id != y.id, SIGN(x.id-y.id), y.val-x.val)))) IN (ROUND((COUNT(*))/2), ROUND((COUNT(*)+1)/2))
) sq
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)
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;
你也可以选择在存储过程中这样做:
DROP PROCEDURE IF EXISTS median;
DELIMITER //
CREATE PROCEDURE median (table_name VARCHAR(255), column_name VARCHAR(255), where_clause VARCHAR(255))
BEGIN
-- Set default parameters
IF where_clause IS NULL OR where_clause = '' THEN
SET where_clause = 1;
END IF;
-- Prepare statement
SET @sql = CONCAT(
"SELECT AVG(middle_values) AS 'median' FROM (
SELECT t1.", column_name, " AS 'middle_values' FROM
(
SELECT @row:=@row+1 as `row`, x.", column_name, "
FROM ", table_name," AS x, (SELECT @row:=0) AS r
WHERE ", where_clause, " ORDER BY x.", column_name, "
) AS t1,
(
SELECT COUNT(*) as 'count'
FROM ", table_name, " x
WHERE ", where_clause, "
) 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
");
-- Execute statement
PREPARE stmt FROM @sql;
EXECUTE stmt;
END//
DELIMITER ;
-- Sample usage:
-- median(table_name, column_name, where_condition);
CALL median('products', 'price', NULL);
