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>

我有下面的代码,我在HackerRank上找到的,它很简单,适用于每一种情况。

SELECT M.MEDIAN_COL FROM MEDIAN_TABLE M WHERE  
  (SELECT COUNT(MEDIAN_COL) FROM MEDIAN_TABLE WHERE MEDIAN_COL < M.MEDIAN_COL ) = 
  (SELECT COUNT(MEDIAN_COL) FROM MEDIAN_TABLE WHERE MEDIAN_COL > M.MEDIAN_COL );

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

你可以使用这里的用户定义函数。

MySQL从8.0版本开始支持窗口函数,您可以使用ROW_NUMBER或DENSE_RANK(不要使用RANK,因为它将相同的RANK分配给相同的值,就像在体育排名):

SELECT AVG(t1.val) AS median_val
  FROM (SELECT val, 
               ROW_NUMBER() OVER(ORDER BY val) AS rownum
          FROM data) t1,
       (SELECT COUNT(*) AS num_records FROM data) t2
 WHERE t1.row_num IN
       (FLOOR((t2.num_records + 1) / 2), 
        FLOOR((t2.num_records + 2) / 2));