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
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。
其他回答
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:-)
我有一个包含大约10亿行的数据库，我们需要它来确定集合中的年龄中位数。对十亿行进行排序是困难的，但如果你将可以找到的不同值(年龄范围从0到100)聚合在一起，你可以对这个列表进行排序，并使用一些算术魔术来找到你想要的任何百分位数，如下所示:
with rawData(count_value) as
(
select p.YEAR_OF_BIRTH
from dbo.PERSON p
),
overallStats (avg_value, stdev_value, min_value, max_value, total) as
(
select avg(1.0 * count_value) as avg_value,
stdev(count_value) as stdev_value,
min(count_value) as min_value,
max(count_value) as max_value,
count(*) as total
from rawData
),
aggData (count_value, total, accumulated) as
(
select count_value,
count(*) as total,
SUM(count(*)) OVER (ORDER BY count_value ROWS UNBOUNDED PRECEDING) as accumulated
FROM rawData
group by count_value
)
select o.total as count_value,
o.min_value,
o.max_value,
o.avg_value,
o.stdev_value,
MIN(case when d.accumulated >= .50 * o.total then count_value else o.max_value end) as median_value,
MIN(case when d.accumulated >= .10 * o.total then count_value else o.max_value end) as p10_value,
MIN(case when d.accumulated >= .25 * o.total then count_value else o.max_value end) as p25_value,
MIN(case when d.accumulated >= .75 * o.total then count_value else o.max_value end) as p75_value,
MIN(case when d.accumulated >= .90 * o.total then count_value else o.max_value end) as p90_value
from aggData d
cross apply overallStats o
GROUP BY o.total, o.min_value, o.max_value, o.avg_value, o.stdev_value
;
这个查询取决于你的db支持窗口函数(包括ROWS UNBOUNDED precede)，但如果你没有，这是一个简单的事情，将aggData CTE与自身连接，并将所有先前的总数聚合到' cumulative '列，用于确定哪个值包含指定的预分词。上面的示例计算p10、p25、p50(中位数)、p75和p90。
屁股的
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
这种方法似乎包括偶数和奇数计数，没有子查询。
SELECT AVG(t1.x)
FROM table t1, table t2
GROUP BY t1.x
HAVING SUM(SIGN(t1.x - t2.x)) = 0
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
