计算按列分组的模式

Question

+--------+-------+
| client | price |
+--------+-------+
|     54 |    25 |
|    648 |    35 |
|     54 |    10 |
|    648 |     8 |
|     54 |    25 |
|    648 |    35 |
+--------+-------+

上面说的是我的 table 模式是如何设置的，我想计算每个 client 最频繁的 price 值，例如。

+--------+-------+
| client | price |
+--------+-------+
|     54 |    25 |
|    648 |    35 |
+--------+-------+

我很难在 MySQL 中完成这项工作。我已经在 PHP 中完成了，就像这样：

$clientPrices = $this->database->select('design', [
    'clientid',
    'price'
]);

$pricesByClients = [];
foreach ($clientPrices as $value) {
    $pricesByClients[$value['clientid']][] = $value['price'];
}

foreach ($pricesByClients as $key => $value) {
    $priceCount = array_count_values($value);
    $mode = array_search(max($priceCount), $priceCount);
    $pricesByClients[$key] = $mode;
}

return $pricesByClients;

但是，这很慢，我希望我可以提高效率或者在 SQL 中完成。

编辑：是 MySQL 5.* 而不是 8.

Answer 1

不幸的是，MySQL 没有内置函数来计算 mode()。

如果您使用的是 MySQL 8.0，则可以使用 window 函数和聚合：

select client, price
from (
    select client, price, rank() over(partition by client order by count(*) desc) rn
    from mytable
    group by client, price
) t
where rn = 1

在早期版本中，选项是使用 having 子句和相关子查询进行过滤

select client, price
from mytable t
group by client, price
having count(*) = (
    select count(*)
    from mytable t1
    where t1.client = t.client
    group by t1.price
    order by count(*) desc
    limit 1
)

Answer 2

对于 MySql 8.0+，您可以使用 row_number() window 函数：

select t.client, t.price
from (
  select client, price,
    row_number() over (partition by client order by count(*) desc) rn
  from tablename
  group by client, price
) t
where t.rn = 1;

对于以前的版本，您可以使用相关子查询：

select distinct t.client, t.price
from tablename t
where (t.client, t.price) = (
  select client, price
  from tablename
  where client = t.client
  group by client, price
  order by count(*) desc
  limit 1
);

参见demo。

计算按列分组的模式

Calculate mode grouped by a column

php

mysql

group-by

count

greatest-n-per-group