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
MySQL从8.0版本开始支持窗口函数，您可以使用ROW_NUMBER或DENSE_RANK(不要使用RANK，因为它将相同的RANK分配给相同的值，就像在体育排名):
SELECT AVG(t1.val) AS median_val
FROM (SELECT val,
ROW_NUMBER() OVER(ORDER BY val) AS rownum
FROM data) t1,
(SELECT COUNT(*) AS num_records FROM data) t2
WHERE t1.row_num IN
(FLOOR((t2.num_records + 1) / 2),
FLOOR((t2.num_records + 2) / 2));
其他回答
MariaDB / MySQL:
SELECT AVG(dd.val) as median_val
FROM (
SELECT d.val, @rownum:=@rownum+1 as `row_number`, @total_rows:=@rownum
FROM data d, (SELECT @rownum:=0) r
WHERE d.val is NOT NULL
-- put some where clause here
ORDER BY d.val
) as dd
WHERE dd.row_number IN ( FLOOR((@total_rows+1)/2), FLOOR((@total_rows+2)/2) );
Steve Cohen指出，在第一次传递之后，@rownum将包含总行数。这可用于确定中值，因此不需要第二次传递或连接。
此外，AVG(dd.val)和dd.row_number IN(…)用于在有偶数条记录时正确地产生中位数。推理:
SELECT FLOOR((3+1)/2),FLOOR((3+2)/2); -- when total_rows is 3, avg rows 2 and 2
SELECT FLOOR((4+1)/2),FLOOR((4+2)/2); -- when total_rows is 4, avg rows 2 and 3
最后，MariaDB 10.3.3+包含一个MEDIAN函数
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。
归档完美中位数的单个查询:
SELECT
COUNT(*) as total_rows,
IF(count(*)%2 = 1, CAST(SUBSTRING_INDEX(SUBSTRING_INDEX( GROUP_CONCAT(val ORDER BY val SEPARATOR ','), ',', 50/100 * COUNT(*)), ',', -1) AS DECIMAL), ROUND((CAST(SUBSTRING_INDEX(SUBSTRING_INDEX( GROUP_CONCAT(val ORDER BY val SEPARATOR ','), ',', 50/100 * COUNT(*) + 1), ',', -1) AS DECIMAL) + CAST(SUBSTRING_INDEX(SUBSTRING_INDEX( GROUP_CONCAT(val ORDER BY val SEPARATOR ','), ',', 50/100 * COUNT(*)), ',', -1) AS DECIMAL)) / 2)) as median,
AVG(val) as average
FROM
data
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));
根据魔术贴的答案，对于那些必须根据另一个参数分组的东西做中位数的人来说
SELECT grp_field, t1。val FROM ( SELECT grp_field， @rownum:=IF(@s = grp_field， @rownum + 1,0) AS row_number， @s:=IF(@s = grp_field， @s, grp_field) AS sec, d.val FROM data d， (SELECT @rownum:=0， @s:=0 ORDER BY grp_field, d.val )作为t1 JOIN ( SELECT grp_field, count(*)为total_rows 数据d GROUP BY grp_field )为t2 在t1。Grp_field = t2.grp_field 在t1.row_number =地板(total_rows / 2) + 1;
