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
下面的SQL代码将帮助您使用用户定义的变量来计算MySQL中的中位数。
create table employees(salary int); insert into employees values(8); insert into employees values(23); insert into employees values(45); insert into employees values(123); insert into employees values(93); insert into employees values(2342); insert into employees values(2238); select * from employees; Select salary from employees order by salary; set @rowid=0; set @cnt=(select count(*) from employees); set @middle_no=ceil(@cnt/2); set @odd_even=null; select AVG(salary) from (select salary,@rowid:=@rowid+1 as rid, (CASE WHEN(mod(@cnt,2)=0) THEN @odd_even:=1 ELSE @odd_even:=0 END) as odd_even_status from employees order by salary) as tbl where tbl.rid=@middle_no or tbl.rid=(@middle_no+@odd_even);
如果你正在寻找详细的解释，请参考这个博客。
其他回答
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;
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;
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;
因为我只需要一个中位数和百分位数的解决方案，我根据这个线程中的发现做了一个简单而相当灵活的函数。我知道，如果我发现“现成的”功能很容易包含在我的项目中，我自己会很高兴，所以我决定快速分享:
function mysql_percentile($table, $column, $where, $percentile = 0.5) {
$sql = "
SELECT `t1`.`".$column."` as `percentile` FROM (
SELECT @rownum:=@rownum+1 as `row_number`, `d`.`".$column."`
FROM `".$table."` `d`, (SELECT @rownum:=0) `r`
".$where."
ORDER BY `d`.`".$column."`
) as `t1`,
(
SELECT count(*) as `total_rows`
FROM `".$table."` `d`
".$where."
) as `t2`
WHERE 1
AND `t1`.`row_number`=floor(`total_rows` * ".$percentile.")+1;
";
$result = sql($sql, 1);
if (!empty($result)) {
return $result['percentile'];
} else {
return 0;
}
}
使用非常简单，例子来自我目前的项目:
...
$table = DBPRE."zip_".$slug;
$column = 'seconds';
$where = "WHERE `reached` = '1' AND `time` >= '".$start_time."'";
$reaching['median'] = mysql_percentile($table, $column, $where, 0.5);
$reaching['percentile25'] = mysql_percentile($table, $column, $where, 0.25);
$reaching['percentile75'] = mysql_percentile($table, $column, $where, 0.75);
...
我刚刚在网上的评论中找到了另一个答案:
对于几乎所有SQL中的中位数: SELECT x.val from data x, data y GROUP BY x.val 总和(符号(1-SIGN (y.val-x.val))) = (COUNT (*) + 1) / 2
确保列有良好的索引，并且索引用于筛选和排序。与解释计划核对。
select count(*) from table --find the number of rows
计算“中值”行号。可能使用:median_row = floor(count / 2)。
然后把它从列表中挑出来:
select val from table order by val asc limit median_row,1
这将返回您想要的值的一行。
