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
我发现接受的解决方案在我的MySQL安装上不起作用，返回一个空集，但这个查询在我测试的所有情况下都适用:
SELECT x.val from data x, data y
GROUP BY x.val
HAVING SUM(SIGN(1-SIGN(y.val-x.val)))/COUNT(*) > .5
LIMIT 1
其他回答
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
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);
SELECT
SUBSTRING_INDEX(
SUBSTRING_INDEX(
GROUP_CONCAT(field ORDER BY field),
',',
((
ROUND(
LENGTH(GROUP_CONCAT(field)) -
LENGTH(
REPLACE(
GROUP_CONCAT(field),
',',
''
)
)
) / 2) + 1
)),
',',
-1
)
FROM
table
上面的方法似乎对我有用。
一个简单的方法来计算中位数在MySQL
set @ct := (select count(1) from station);
set @row := 0;
select avg(a.val) as median from
(select * from table order by val) a
where (select @row := @row + 1)
between @ct/2.0 and @ct/2.0 +1;
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)
