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
按维度分组的中位数:
SELECT your_dimension, avg(t1.val) as median_val FROM (
SELECT @rownum:=@rownum+1 AS `row_number`,
IF(@dim <> d.your_dimension, @rownum := 0, NULL),
@dim := d.your_dimension AS your_dimension,
d.val
FROM data d, (SELECT @rownum:=0) r, (SELECT @dim := 'something_unreal') d
WHERE 1
-- put some where clause here
ORDER BY d.your_dimension, d.val
) as t1
INNER JOIN
(
SELECT d.your_dimension,
count(*) as total_rows
FROM data d
WHERE 1
-- put same where clause here
GROUP BY d.your_dimension
) as t2 USING(your_dimension)
WHERE 1
AND t1.row_number in ( floor((total_rows+1)/2), floor((total_rows+2)/2) )
GROUP BY your_dimension;
其他回答
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)
上面的大多数解决方案只适用于表中的一个字段，您可能需要获得查询中多个字段的中位数(第50百分位数)。
我用这个:
SELECT CAST(SUBSTRING_INDEX(SUBSTRING_INDEX(
GROUP_CONCAT(field_name ORDER BY field_name SEPARATOR ','),
',', 50/100 * COUNT(*) + 1), ',', -1) AS DECIMAL) AS `Median`
FROM table_name;
你可以将上面例子中的“50”替换为任何百分位数，这是非常有效的。
只要确保你有足够的内存给GROUP_CONCAT，你可以改变它:
SET group_concat_max_len = 10485760; #10MB max length
详情:http://web.performancerasta.com/metrics-tips-calculating-95th-99th-or-any-percentile-with-single-mysql-query/
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;
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));
让我们创建一个名为numbers的示例表
这个答案是针对mysql数据库的
在postgres Sql中，它简单地使用per_cont函数
创建表数字( num INT, 频率整数 ）;
在数字表中插入值
插入数字 (7) 0 (1, 1), (2、3), (1) 3 (9,1), (1, 1), (2、3), (1) 3 (9,1);
——select * from numbers
作为递归num_frequency (num,frequency, i) （ 选择num,频率,1 从数字 UNION ALL 选择num,频率,i + 1 从num_frequency num_frequency的地方。I < num_frequency.frequency ）
select * (max(当numbers=lower_limit时，则num else null end)/2 +max(当数字=upper_limit时，则num else null end)/2)作为中位数 从( select *, total_number % 2, 情况下 当total_number%2=0时，total_number/2 Else (total_number+1)/2 end as lower_limit， 情况下 当total_number%2=0时，total_number/2+1 其他(total_number + 1) / 2 结束为upper_limit
从( Select *，max(numbers) over() as total_number from ( Select num,row_number() over(按num排序) 作为num_frequency中的数字 b) b) b)
