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

我使用下表的解决方案在MySQL:

CREATE TABLE transactions (
  transaction_id int , user_id int , merchant_name varchar(255), transaction_date date , amount int
);

INSERT INTO transactions (transaction_id, user_id, merchant_name, transaction_date, amount)  
VALUES (1, 1 ,'abc', '2015-08-17', 100),(2, 2, 'ced', '2015-2-17', 100),(3, 1, 'def', '2015-2-16', 121),
(4, 1 ,'ced', '2015-3-17', 110),(5, 1, 'ced', '2015-3-17', 150),(6, 2 ,'abc', '2015-4-17', 130), 
(7, 3 ,'ced', '2015-12-17', 10),(8, 3 ,'abc', '2015-8-17', 100),(9, 2 ,'abc', '2015-12-17', 140),(10, 1,'abc', '2015-9-17', 100),
(11, 1 ,'abc', '2015-08-17', 121),(12, 2 ,'ced', '2015-12-23', 130),(13, 1 ,'def', '2015-12-23', 13),(3, 4, 'abc', '2015-2-16', 120),(3, 4, 'def', '2015-2-16', 121),(3, 4, 'ced', '2015-2-16', 121);

计算“金额”列的中位数:

WITH Numbered AS 
(
SELECT *, COUNT(*) OVER () AS TotatRecords,
    ROW_NUMBER() OVER (ORDER BY amount) AS RowNum
FROM transactions
)
SELECT Avg(amount)
FROM Numbered
WHERE RowNum IN ( FLOOR((TotatRecords+1)/2), FLOOR((TotatRecords+2)/2) )
;

TotalRecords = 16 and Median = 120.5000

此查询将适用于两种情况,即偶数和奇数记录。

其他回答

来自: http://mdb-blog.blogspot.com/2015/06/mysql-find-median-nth-element-without.html

我建议另一种方法,不用连接, 但是处理字符串

我没有用大数据表来检验, 但是中小型的桌子就可以了。

这里的好处是,它也可以通过GROUPING工作,所以它可以返回几个项目的中值。

下面是测试表的测试代码:

DROP TABLE test.test_median
CREATE TABLE test.test_median AS
SELECT 'book' AS grp, 4 AS val UNION ALL
SELECT 'book', 7 UNION ALL
SELECT 'book', 2 UNION ALL
SELECT 'book', 2 UNION ALL
SELECT 'book', 9 UNION ALL
SELECT 'book', 8 UNION ALL
SELECT 'book', 3 UNION ALL

SELECT 'note', 11 UNION ALL

SELECT 'bike', 22 UNION ALL
SELECT 'bike', 26 

求每组中位数的代码:

SELECT grp,
         SUBSTRING_INDEX( SUBSTRING_INDEX( GROUP_CONCAT(val ORDER BY val), ',', COUNT(*)/2 ), ',', -1) as the_median,
         GROUP_CONCAT(val ORDER BY val) as all_vals_for_debug
FROM test.test_median
GROUP BY grp

输出:

grp | the_median| all_vals_for_debug
bike| 22        | 22,26
book| 4         | 2,2,3,4,7,8,9
note| 11        | 11

基于@bob的回答,这将查询泛化为能够返回多个中位数,并按某些标准分组。

想想,例如,一个车场二手车的中位数销售价格,按年-月分组。

SELECT 
    period, 
    AVG(middle_values) AS 'median' 
FROM (
    SELECT t1.sale_price AS 'middle_values', t1.row_num, t1.period, t2.count
    FROM (
        SELECT 
            @last_period:=@period AS 'last_period',
            @period:=DATE_FORMAT(sale_date, '%Y-%m') AS 'period',
            IF (@period<>@last_period, @row:=1, @row:=@row+1) as `row_num`, 
            x.sale_price
          FROM listings AS x, (SELECT @row:=0) AS r
          WHERE 1
            -- where criteria goes here
          ORDER BY DATE_FORMAT(sale_date, '%Y%m'), x.sale_price
        ) AS t1
    LEFT JOIN (  
          SELECT COUNT(*) as 'count', DATE_FORMAT(sale_date, '%Y-%m') AS 'period'
          FROM listings x
          WHERE 1
            -- same where criteria goes here
          GROUP BY DATE_FORMAT(sale_date, '%Y%m')
        ) AS t2
        ON t1.period = t2.period
    ) AS t3
WHERE 
    row_num >= (count/2) 
    AND row_num <= ((count/2) + 1)
GROUP BY t3.period
ORDER BY t3.period;

你也可以选择在存储过程中这样做:

DROP PROCEDURE IF EXISTS median;
DELIMITER //
CREATE PROCEDURE median (table_name VARCHAR(255), column_name VARCHAR(255), where_clause VARCHAR(255))
BEGIN
  -- Set default parameters
  IF where_clause IS NULL OR where_clause = '' THEN
    SET where_clause = 1;
  END IF;

  -- Prepare statement
  SET @sql = CONCAT(
    "SELECT AVG(middle_values) AS 'median' FROM (
      SELECT t1.", column_name, " AS 'middle_values' FROM
        (
          SELECT @row:=@row+1 as `row`, x.", column_name, "
          FROM ", table_name," AS x, (SELECT @row:=0) AS r
          WHERE ", where_clause, " ORDER BY x.", column_name, "
        ) AS t1,
        (
          SELECT COUNT(*) as 'count'
          FROM ", table_name, " x
          WHERE ", where_clause, "
        ) 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
    ");

  -- Execute statement
  PREPARE stmt FROM @sql;
  EXECUTE stmt;
END//
DELIMITER ;


-- Sample usage:
-- median(table_name, column_name, where_condition);
CALL median('products', 'price', NULL);

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。

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;