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
不幸的是，无论是TheJacobTaylor还是velcrow的答案都不会返回当前版本MySQL的准确结果。
从上面来看，魔术贴的答案是接近的，但它不能正确计算具有偶数行数的结果集。中位数定义为1)奇数集上的中间数，或2)偶数集上两个中间数的平均值。
所以，这里是魔术贴的解决方案修补处理奇数和偶数集:
SELECT AVG(middle_values) AS 'median' FROM (
SELECT t1.median_column AS 'middle_values' FROM
(
SELECT @row:=@row+1 as `row`, x.median_column
FROM median_table AS x, (SELECT @row:=0) AS r
WHERE 1
-- put some where clause here
ORDER BY x.median_column
) AS t1,
(
SELECT COUNT(*) as 'count'
FROM median_table x
WHERE 1
-- put same where clause here
) 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;
要使用它，请遵循以下3个简单步骤:
将上面代码中的“median_table”(出现2次)替换为您的表名 将“median_column”(3次)替换为您希望为其查找中位数的列名 如果你有一个WHERE条件，用WHERE条件替换“WHERE 1”(2次)
另一个对Velcrow答案的重复，但使用了一个中间表，并利用了用于行编号的变量来获得计数，而不是执行额外的查询来计算它。还开始计数，以便第一行是第0行，以便简单地使用Floor和Ceil选择中位数行。
SELECT Avg(tmp.val) as median_val
FROM (SELECT inTab.val, @rows := @rows + 1 as rowNum
FROM data as inTab, (SELECT @rows := -1) as init
-- Replace with better where clause or delete
WHERE 2 > 1
ORDER BY inTab.val) as tmp
WHERE tmp.rowNum in (Floor(@rows / 2), Ceil(@rows / 2));
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
因为我只需要一个中位数和百分位数的解决方案，我根据这个线程中的发现做了一个简单而相当灵活的函数。我知道，如果我发现“现成的”功能很容易包含在我的项目中，我自己会很高兴，所以我决定快速分享:
function mysql_percentile($table, $column, $where, $percentile = 0.5) {
$sql = "
SELECT `t1`.`".$column."` as `percentile` FROM (
SELECT @rownum:=@rownum+1 as `row_number`, `d`.`".$column."`
FROM `".$table."` `d`, (SELECT @rownum:=0) `r`
".$where."
ORDER BY `d`.`".$column."`
) as `t1`,
(
SELECT count(*) as `total_rows`
FROM `".$table."` `d`
".$where."
) as `t2`
WHERE 1
AND `t1`.`row_number`=floor(`total_rows` * ".$percentile.")+1;
";
$result = sql($sql, 1);
if (!empty($result)) {
return $result['percentile'];
} else {
return 0;
}
}
使用非常简单，例子来自我目前的项目:
...
$table = DBPRE."zip_".$slug;
$column = 'seconds';
$where = "WHERE `reached` = '1' AND `time` >= '".$start_time."'";
$reaching['median'] = mysql_percentile($table, $column, $where, 0.5);
$reaching['percentile25'] = mysql_percentile($table, $column, $where, 0.25);
$reaching['percentile75'] = mysql_percentile($table, $column, $where, 0.75);
...
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));
