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
通常，我们不仅需要为整个表计算Median，还需要为与ID相关的聚合计算Median。换句话说，计算表中每个ID的中位数，其中每个ID有许多记录。(良好的性能和工作在许多SQL +修复偶数和赔率的问题，更多关于不同的中值方法的性能https://sqlperformance.com/2012/08/t-sql-queries/median)
SELECT our_id, AVG(1.0 * our_val) as Median
FROM
( SELECT our_id, our_val,
COUNT(*) OVER (PARTITION BY our_id) AS cnt,
ROW_NUMBER() OVER (PARTITION BY our_id ORDER BY our_val) AS rn
FROM our_table
) AS x
WHERE rn IN ((cnt + 1)/2, (cnt + 2)/2) GROUP BY our_id;
希望能有所帮助
其他回答
我发现这个答案非常有用——https://www.eversql.com/how-to-calculate-median-value-in-mysql-using-a-simple-sql-query/
SET @rowindex := -1;
SELECT
AVG(g.grade)
FROM
(SELECT @rowindex:=@rowindex + 1 AS rowindex,
grades.grade AS grade
FROM grades
ORDER BY grades.grade) AS g
WHERE
g.rowindex IN (FLOOR(@rowindex / 2) , CEIL(@rowindex / 2));
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;
对于一个表站和列lat_n，下面是MySQL代码来获得中位数:
set @rows := (select count(1) from station);
set @v1 := 0;
set @sql1 := concat('select lat_n into @v1 from station order by lat_n asc limit 1 offset ', ceil(@rows/2) - 1);
prepare statement1 from @sql1;
execute statement1;
set @v2 := 0;
set @sql2 := concat('select lat_n into @v2 from station order by lat_n asc limit 1 offset ', ceil((@rows + 1)/2) - 1);
prepare statement2 from @sql2;
execute statement2;
select (@v1 + @v2)/2;
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)
