我有一个名为 offers 的相当大的数据集,其中包含大约 700 万行。
该表有 30 列,但我只使用其中的两列,cap_id - 车辆的唯一标识符,以及 price - 每月租赁成本车辆。
我想编写一个查询,返回每个 cap_id 的最佳(最低)和次佳价格,以及与次佳价格相比,最佳价格节省的百分比。
我使用的是 5.7.12 版
创建表查询:
closePrices <- prices[Close]
这是我迄今为止尝试过的:
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);
这将返回 0 行,并且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 | 选择类型 | 表 | 分区 | 类型 | possible_keys | 键 | key_len | 参考 | 行 | 过滤 | 额外 | FIELD13 | FIELD14 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|
1 | 简单 | x | 索引 | cap_id | idx_profile_grouping | idx_capId_monthlyPayment | idx_capId_monthlyPayment | 9 | 7220930 | 100.00 | 使用索引;使用临时;使用文件排序 | ||
1 | 简单 | y | 参考 | cap_id | idx_profile_grouping | idx_capId_monthlyPayment | idx_capId_monthlyPayment | 4 | moneyshake.x.cap_id | 871 | 33.33 | 使用哪里;使用索引 |
预先感谢您的任何建议。
答案 0 :(得分:1)
可以在较新版本的 MySQL 中优化此过程,但由于您的小提琴是 5.6,所以我的答案是:
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;
答案 1 :(得分:1)
在 MySQL 8.x 中你可以:
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