MySQL - 按 ID 选择第二个最小值

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【中文标题】MySQL - 按 ID 选择第二个最小值【英文标题】:MySQL - select 2nd lowest value by ID 【发布时间】:2021-07-23 17:54:16 【问题描述】:

我有一个相当大的数据集,名为 offers,包含大约 7m 行。

该表有 30 列,但我只使用其中两列,cap_id - 车辆的唯一标识符,price - 每月租赁费用车辆。

我想编写一个查询,返回每个 cap_id 的最佳(最低)和次佳价格,以及最佳价格与次佳价格相比节省的百分比。

我使用的是 5.7.12 版

这是SQLFiddle

创建表查询:

CREATE TABLE `offers` (
  `id` mediumint(8) unsigned NOT NULL auto_increment,
  `cap_id` varchar(255) default NULL,
  `price` mediumint default NULL,
  PRIMARY KEY (`id`)
) AUTO_INCREMENT=1;

INSERT INTO `offers` (`cap_id`,`price`) VALUES 
(18452,1007),(18452,884),(18452,276),(90019,328),(73353,539),(64854,249),(26684,257),(37452,966),(90019,980),(73353,1241),
(73353,1056),(37452,1043),(26684,829),(37452,260),(64854,358),(26684,288),(26684,678),(26684,905),(37452,1140),(94826,901),
(90019,745),(37452,1156),(37452,191),(64854,324),(73353,1110),(87725,624),(87725,973),(90019,1203),(90019,709),(18452,1133),
(18452,1019),(37452,639),(37452,1021),(87725,485),(94826,964),(37452,1066),(94826,823),(73353,1056),(18452,621),(37452,272),
(90019,223),(26684,412),(87725,310),(37452,948),(37452,826),(18452,1078),(90019,737),(18452,1166),(73353,150),(73353,1115),
(94826,957),(87725,242),(94826,715),(73353,1190),(94826,320),(94826,869),(64854,574),(94826,505),(26684,322),(90019,949),
(64854,1188),(37452,368),(90019,796),(87725,514),(37452,146),(94826,1216),(18452,625),(64854,1165),(18452,712),(37452,947),
(64854,616),(73353,1065),(26684,1167),(18452,935),(87725,1192),(26684,519),(64854,939),(90019,367),(26684,145),(64854,1076),
(26684,1016),(90019,606),(37452,1066),(73353,609),(94826,343),(94826,236),(94826,1059),(26684,681),(37452,779),(94826,259),
(87725,1080),(37452,914),(90019,826),(37452,597),(26684,879),(87725,471),(94826,680),(18452,906),(87725,860),(94826,1009);

这是我迄今为止尝试过的:

SELECT 
 o1.cap_id,
 o2.price AS best_price,
 o1.price AS next_best,
 (o1.price / o2.price) * 100 AS '%_diff'
FROM
 offers o1
     JOIN
 offers o2 ON o1.cap_id = o2.cap_id
     AND o1.price > o2.price
GROUP BY o1.cap_id
HAVING COUNT(o1.price) = 2

这会返回 0 行,当我在我们的数据库中运行它时运行速度非常慢。

这是 EXPLAIN 的输出:

id select_type table partitions type possible_keys key key_len ref rows filtered Extra FIELD13 FIELD14
1 SIMPLE x index cap_id idx_profile_grouping idx_capId_monthlyPayment idx_capId_monthlyPayment 9 7220930 100.00 Using index; Using temporary; Using filesort
1 SIMPLE y ref cap_id idx_profile_grouping idx_capId_monthlyPayment idx_capId_monthlyPayment 4 moneyshake.x.cap_id 871 33.33 Using where; Using index

提前感谢您的任何建议。

【问题讨论】:

您链接到 5.6 版。是你用的那个版本吗? 这在 mysql 8 中很容易。在旧版本中这有点尴尬。所以,请回答草莓的问题。你用的是哪个版本? @Strawberry 刚刚编辑 - 使用 5.7.12,谢谢 如果一个 cap_id 的最优价格出现两次,那么最优价格是否等于次优价格?还是您正在寻找下一个更低的价格? 5.7 是一个相当老的版本。你不能升级到 MySQL 8 吗?它提供了 MySQL 一直缺乏的许多特性(窗口函数、递归和非递归 CTE)。 【参考方案1】:

这个过程可以在较新版本的 MySQL 中优化,但是你的小提琴在 5.6 中,所以我的回答是:

SELECT x.*
     , COUNT(*) running 
  FROM offers x 
  JOIN offers y 
    ON y.cap_id = x.cap_id 
   AND y.price < x.price 
 GROUP 
    BY x.id
 ORDER
    BY x.cap_id 
     , x.price;

扩展这个想法:

SELECT a.*
     , b.price
     , 1-(a.price/b.price) saving
  FROM 
     ( SELECT cap_id
            , MIN(price) price
         FROM offers 
        GROUP  
           BY cap_id
     ) a -- lowest price per cap_id
  JOIN 
     ( SELECT x.cap_id
            , x.price
         FROM offers x
         JOIN offers y
           ON y.cap_id = x.cap_id 
          AND y.price < x.price 
        GROUP 
           BY x.id
       HAVING COUNT(*)= 2
    ) b -- 2nd lowest price per cap_id (other methods are available)
   ON b.cap_id = a.cap_id
ORDER 
   BY a.cap_id;

【讨论】:

您指的是哪个“拼图的最后一块”我不确定-我尝试运行您的代码并且这个:SELECT x.cap_id, y.price as best_price, x.price as next_best FROM offers x JOIN offers y ON y.cap_id = x.cap_id AND y.price &lt; x.price GROUP BY x.cap_id ORDER BY x.cap_id , x.price; 两者都不起作用-它们只是永远运行。 我的代码似乎运行良好:sqlfiddle.com/#!9/ddd9eb/13 我想性能是这里的问题。我也不确定如何返回最优价格与次优相比的“节省”百分比【参考方案2】:

在 MySQL 8.x 中你可以这样做:

with
p as (
  select
    id, cap_id, price,
    row_number() over(partition by cap_id order by price) as rn
  from offers
)
select 
  a.id as lowest_id, a.cap_id as lowest_cap_id, a.price as lowest_price,
  b.id as second_id, b.cap_id as second_cap_id, b.price as second_price,
  case when b.price is not null then 
    (b.price - a.price) / b.price
  end as percentage_saving
from p a
left join p b on a.cap_id = b.cap_id and b.rn = 2
where a.rn = 1

【讨论】:

在 MySQL 工作台中,它说“目标 mysql 版本”是 8.XX,但我们的 RDS 实例仍在运行 5.7。我猜是时候考虑升级了。

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