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

我没有将这个解决方案的性能与这里发布的其他答案进行比较,但我发现这个解决方案是最容易理解的,并且涵盖了计算中位数的全部数学公式。换句话说,这个解决方案对于偶数和奇数数据集足够健壮:

SELECT CASE 
-- odd-numbered data sets:
WHEN MOD(COUNT(*), 2) = 1 THEN (SELECT median.<value> AS median
FROM
(SELECT t1.<value>
  FROM (SELECT <value>, 
               ROW_NUMBER() OVER(ORDER BY <value>) AS rownum
          FROM <data>) t1,
       (SELECT COUNT(*) AS num_records FROM <data>) t2
 WHERE t1.rownum =(t2.num_records) / 2) as median)
-- even-numbered data sets:
ELSE (select (low_bound.<value> + up_bound.<value>) / 2 AS median
FROM
(SELECT t1.<value>
  FROM (SELECT <value>, 
               ROW_NUMBER() OVER(ORDER BY <value>) AS rownum
          FROM <data>) t1,
       (SELECT COUNT(*) AS num_records FROM <data>) t2
 WHERE t1.rownum =(t2.num_records - 1) / 2) as low_bound,
 (SELECT t1.<value>
  FROM (SELECT <value>, 
               ROW_NUMBER() OVER(ORDER BY <value>) AS rownum
          FROM station) t1,
       (SELECT COUNT(*) AS num_records FROM data) t2
 WHERE t1.rownum =(t2.num_records + 1) / 2) as up_bound)
END
FROM <data>

其他回答

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

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;

我使用了两个查询方法:

第一个得到count, min, Max和avg 第二个语句(预处理语句)使用“LIMIT @count/ 2,1”和“ORDER BY ..”子句来获得中值

它们被包装在函数defn中,因此可以从一次调用中返回所有值。

如果您的范围是静态的,并且数据不经常更改,那么预先计算/存储这些值并使用存储的值,而不是每次都从头查询,可能会更有效。

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);
...

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));