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
试着这样做:
SELECT
CAST (AVG(val) AS DECIMAL(10,4))
FROM
(
SELECT
val,
ROW_NUMBER() OVER( ORDER BY val ) -1 AS rn,
COUNT(1) OVER () -1 AS cnt
FROM STATION
) as tmp
WHERE rn IN (FLOOR(cnt/2),CEILING (cnt/2))
**
注意:-1的原因是使它的索引为0 .i。E行号 现在从0开始，而不是1
**
其他回答
最简单和快速的方法来计算中位数在mysql。
select x.col
from (select lat_n,
count(1) over (partition by 'A') as total_rows,
row_number() over (order by col asc) as rank_Order
from station ft) x
where x.rank_Order = round(x.total_rows / 2.0, 0)
我的代码，高效，没有表或额外的变量:
SELECT
((SUBSTRING_INDEX(SUBSTRING_INDEX(group_concat(val order by val), ',', floor(1+((count(val)-1) / 2))), ',', -1))
+
(SUBSTRING_INDEX(SUBSTRING_INDEX(group_concat(val order by val), ',', ceiling(1+((count(val)-1) / 2))), ',', -1)))/2
as median
FROM table;
在某些情况下，中位数的计算如下:
“中位数”是数字列表中按值排序时的“中间”值。对于偶数集，中位数是两个中间值的平均值。 我为此创建了一个简单的代码:
$midValue = 0;
$rowCount = "SELECT count(*) as count {$from} {$where}";
$even = FALSE;
$offset = 1;
$medianRow = floor($rowCount / 2);
if ($rowCount % 2 == 0 && !empty($medianRow)) {
$even = TRUE;
$offset++;
$medianRow--;
}
$medianValue = "SELECT column as median
{$fromClause} {$whereClause}
ORDER BY median
LIMIT {$medianRow},{$offset}";
$medianValDAO = db_query($medianValue);
while ($medianValDAO->fetch()) {
if ($even) {
$midValue = $midValue + $medianValDAO->median;
}
else {
$median = $medianValDAO->median;
}
}
if ($even) {
$median = $midValue / 2;
}
return $median;
返回的$中位数将是所需的结果:-)
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;
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
