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
知道确切的行数，你可以使用这个查询:
SELECT <value> AS VAL FROM <table> ORDER BY VAL LIMIT 1 OFFSET <half>
Where <half> = ceiling(<size> / 2.0) - 1
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);
...
我没有将这个解决方案的性能与这里发布的其他答案进行比较，但我发现这个解决方案是最容易理解的，并且涵盖了计算中位数的全部数学公式。换句话说，这个解决方案对于偶数和奇数数据集足够健壮:
SELECT CASE
-- odd-numbered data sets:
WHEN MOD(COUNT(*), 2) = 1 THEN (SELECT median.<value> AS median
FROM
(SELECT t1.<value>
FROM (SELECT <value>,
ROW_NUMBER() OVER(ORDER BY <value>) AS rownum
FROM <data>) t1,
(SELECT COUNT(*) AS num_records FROM <data>) t2
WHERE t1.rownum =(t2.num_records) / 2) as median)
-- even-numbered data sets:
ELSE (select (low_bound.<value> + up_bound.<value>) / 2 AS median
FROM
(SELECT t1.<value>
FROM (SELECT <value>,
ROW_NUMBER() OVER(ORDER BY <value>) AS rownum
FROM <data>) t1,
(SELECT COUNT(*) AS num_records FROM <data>) t2
WHERE t1.rownum =(t2.num_records - 1) / 2) as low_bound,
(SELECT t1.<value>
FROM (SELECT <value>,
ROW_NUMBER() OVER(ORDER BY <value>) AS rownum
FROM station) t1,
(SELECT COUNT(*) AS num_records FROM data) t2
WHERE t1.rownum =(t2.num_records + 1) / 2) as up_bound)
END
FROM <data>
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
