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
通常，我们不仅需要为整个表计算Median，还需要为与ID相关的聚合计算Median。换句话说，计算表中每个ID的中位数，其中每个ID有许多记录。(良好的性能和工作在许多SQL +修复偶数和赔率的问题，更多关于不同的中值方法的性能https://sqlperformance.com/2012/08/t-sql-queries/median)
SELECT our_id, AVG(1.0 * our_val) as Median
FROM
( SELECT our_id, our_val,
COUNT(*) OVER (PARTITION BY our_id) AS cnt,
ROW_NUMBER() OVER (PARTITION BY our_id ORDER BY our_val) AS rn
FROM our_table
) AS x
WHERE rn IN ((cnt + 1)/2, (cnt + 2)/2) GROUP BY our_id;
希望能有所帮助
其他回答
set @r = 0;
select
case when mod(c,2)=0 then round(sum(lat_N),4)
else round(sum(lat_N)/2,4)
end as Med
from
(select lat_N, @r := @r+1, @r as id from station order by lat_N) A
cross join
(select (count(1)+1)/2 as c from station) B
where id >= floor(c) and id <=ceil(c)
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。
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)
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;
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);
...
