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。
我下面提出的解决方案只需要一个查询,而不需要创建表、变量甚至子查询。
另外,它可以让你在组-查询(这是我需要的!)
SELECT `columnA`,
SUBSTRING_INDEX(SUBSTRING_INDEX(GROUP_CONCAT(`columnB` ORDER BY `columnB`), ',', CEILING((COUNT(`columnB`)/2))), ',', -1) medianOfColumnB
FROM `tableC`
-- some where clause if you want
GROUP BY `columnA`;
它之所以能够工作,是因为巧妙地使用了group_concat和substring_index。
但是,为了允许大的group_concat,必须将group_concat_max_len设置为一个更高的值(默认为1024字符)。
你可以这样设置(对于当前的sql会话):
SET SESSION group_concat_max_len = 10000;
-- up to 4294967295 in 32-bits platform.
有关group_concat_max_len的更多信息:https://dev.mysql.com/doc/refman/5.1/en/server-system-variables.html#sysvar_group_concat_max_len
MariaDB / MySQL:
SELECT AVG(dd.val) as median_val
FROM (
SELECT d.val, @rownum:=@rownum+1 as `row_number`, @total_rows:=@rownum
FROM data d, (SELECT @rownum:=0) r
WHERE d.val is NOT NULL
-- put some where clause here
ORDER BY d.val
) as dd
WHERE dd.row_number IN ( FLOOR((@total_rows+1)/2), FLOOR((@total_rows+2)/2) );
Steve Cohen指出,在第一次传递之后,@rownum将包含总行数。这可用于确定中值,因此不需要第二次传递或连接。
此外,AVG(dd.val)和dd.row_number IN(…)用于在有偶数条记录时正确地产生中位数。推理:
SELECT FLOOR((3+1)/2),FLOOR((3+2)/2); -- when total_rows is 3, avg rows 2 and 2
SELECT FLOOR((4+1)/2),FLOOR((4+2)/2); -- when total_rows is 4, avg rows 2 and 3
最后,MariaDB 10.3.3+包含一个MEDIAN函数
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);
这些方法从同一个表中选择两次。如果源数据来自一个昂贵的查询,这是一种避免运行两次的方法:
select KEY_FIELD, AVG(VALUE_FIELD) MEDIAN_VALUE
from (
select KEY_FIELD, VALUE_FIELD, RANKF
, @rownumr := IF(@prevrowidr=KEY_FIELD,@rownumr+1,1) RANKR
, @prevrowidr := KEY_FIELD
FROM (
SELECT KEY_FIELD, VALUE_FIELD, RANKF
FROM (
SELECT KEY_FIELD, VALUE_FIELD
, @rownumf := IF(@prevrowidf=KEY_FIELD,@rownumf+1,1) RANKF
, @prevrowidf := KEY_FIELD
FROM (
SELECT KEY_FIELD, VALUE_FIELD
FROM (
-- some expensive query
) B
ORDER BY KEY_FIELD, VALUE_FIELD
) C
, (SELECT @rownumf := 1) t_rownum
, (SELECT @prevrowidf := '*') t_previd
) D
ORDER BY KEY_FIELD, RANKF DESC
) E
, (SELECT @rownumr := 1) t_rownum
, (SELECT @prevrowidr := '*') t_previd
) F
WHERE RANKF-RANKR BETWEEN -1 and 1
GROUP BY KEY_FIELD
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。