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

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

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));

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)

安装和使用本mysql统计函数:http://www.xarg.org/2012/07/statistical-functions-in-mysql/

之后,计算中值就很简单了:

SELECT median(val) FROM data;

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));