行转列-有序行转列

数据仓库晨曦

发布于 2024-08-13 18:30:09

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发布于 2024-08-13 18:30:09

文章被收录于专栏：数据仓库技术

一、基础数据

有配送订单表记录骑手配送的物品类型、送达时间、顾客id、配送举例及配送费。

+-----------+-----------+-------------+----------------------+--------------+-----------+----------+
| rider_id  | order_id  | goods_type  |    delivery_time     | customer_id  | distance  | payment  |
+-----------+-----------+-------------+----------------------+--------------+-----------+----------+
| r001      | 0001      | 食品          | 2024-01-01 08:30:00  | c001         | 8.05      | 7.50     |
| r002      | 0002      | 酒水          | 2024-01-01 08:43:00  | c003         | 3.01      | 13.00    |
| r003      | 0003      | 食品          | 2024-01-01 09:15:00  | c004         | 4.12      | 3.50     |
| r002      | 0004      | 文件          | 2024-01-01 09:21:00  | c005         | 10.98     | 15.00    |
| r001      | 0005      | 食品          | 2024-01-01 09:30:00  | c007         | 2.32      | 5.00     |
| r002      | 0006      | 食品          | 2024-01-01 09:47:00  | c002         | 0.78      | 5.00     |
| r003      | 0007      | 文件          | 2024-01-01 10:21:00  | c010         | 8.11      | 8.00     |
| r001      | 0008      | 酒水          | 2024-01-01 11:56:00  | c023         | 4.35      | 15.00    |
| r002      | 0009      | 电子产品        | 2024-01-01 12:30:00  | c031         | 5.05      | 9.50     |
| r002      | 0010      | 文件          | 2024-01-01 13:30:00  | c001         | 6.05      | 7.00     |
+-----------+-----------+-------------+----------------------+--------------+-----------+----------+

二、函数介绍

1.transform介绍

transform(expr, func) - 使用函数对数组中的元素进行转换。

Examples:

> SELECT transform(array(1, 2, 3), x -> x + 1);
 [2,3,4]
> SELECT transform(array(1, 2, 3), (x, i) -> x + i);
 [1,3,5]

Since: 2.4.0

2.其他函数介绍

collect_list
collect_set
concat_ws

三、有序行转列

根据配送订单记录表，查询出骑手id，顾客id列表，要求顾客id列表中的顺序按照送达时间早晚排序。

分析

这里要求按照订单送达时间，对顾客id进行排序。直接考虑是使用开窗函数，根据时间进行排序（这种方法不可行，原因在于collect_list和collect_set那句提示“该函数是非确定性的，因为收集结果的顺序取决于行的顺序，这在经过shuffle之后可能是不确定的”。collect_list函数很难测试，但是collect_set因为有去重操作，所以比较好测试。另外一种解法是将时间和用户id拼接在一起，时间在前，用户id在后，这样对整个字符串拼接为数组，进行排序后再拆分，如此得到的数据能确定保证按照顺序完成。

1.不确定解法（不建议使用）

collect_list执行SQL

select rider_id,
       delivery_time,
       customer_id,
       collect_list(customer_id)over(partition by rider_id order by delivery_time) as customer_id_list
from t_delivery_orders

collect_list执行结果

+-----------+----------------------+--------------+---------------------------------------+
| rider_id  |    delivery_time     | customer_id  |           customer_id_list            |
+-----------+----------------------+--------------+---------------------------------------+
| r001      | 2024-01-01 08:30:00  | c001         | ["c001"]                              |
| r001      | 2024-01-01 09:30:00  | c007         | ["c001","c007"]                       |
| r001      | 2024-01-01 11:56:00  | c023         | ["c001","c007","c023"]                |
| r002      | 2024-01-01 08:43:00  | c003         | ["c003"]                              |
| r002      | 2024-01-01 09:21:00  | c005         | ["c003","c005"]                       |
| r002      | 2024-01-01 09:47:00  | c002         | ["c003","c005","c002"]                |
| r002      | 2024-01-01 12:30:00  | c031         | ["c003","c005","c002","c031"]         |
| r002      | 2024-01-01 13:30:00  | c001         | ["c003","c005","c002","c031","c001"]  |
| r003      | 2024-01-01 09:15:00  | c004         | ["c004"]                              |
| r003      | 2024-01-01 10:21:00  | c010         | ["c004","c010"]                       |
+-----------+----------------------+--------------+---------------------------------------+

collect_set执行SQL

select rider_id,
       delivery_time,
       customer_id,
       collect_set(customer_id)over(partition by rider_id order by delivery_time) as customer_id_list
from t_delivery_orders

collect_set执行结果

+-----------+----------------------+--------------+---------------------------------------+
| rider_id  |    delivery_time     | customer_id  |           customer_id_list            |
+-----------+----------------------+--------------+---------------------------------------+
| r001      | 2024-01-01 08:30:00  | c001         | ["c001"]                              |
| r001      | 2024-01-01 09:30:00  | c007         | ["c001","c007"]                       |
| r001      | 2024-01-01 11:56:00  | c023         | ["c001","c007","c023"]                |
| r002      | 2024-01-01 08:43:00  | c003         | ["c003"]                              |
| r002      | 2024-01-01 09:21:00  | c005         | ["c005","c003"]                       |
| r002      | 2024-01-01 09:47:00  | c002         | ["c005","c003","c002"]                |
| r002      | 2024-01-01 12:30:00  | c031         | ["c005","c003","c002","c031"]         |
| r002      | 2024-01-01 13:30:00  | c001         | ["c001","c005","c003","c002","c031"]  |
| r003      | 2024-01-01 09:15:00  | c004         | ["c004"]                              |
| r003      | 2024-01-01 10:21:00  | c010         | ["c010","c004"]                       |
+-----------+----------------------+--------------+---------------------------------------+

我们可以观察最后一行结果，发现顺序是不一样。 可以看到collect_list结果是正确的（这里属于巧合，但是大多数情况都是对的），而collect_set的结果是错误的。两个函数中均有提示，不保证结果顺序，从严谨出发，这两个函数均不可因为开窗函数来保证有序。 如果忽略该问题，使用collect_list进行行转列开窗，大部分情况得到的结果是正确的。我们只需要取出每个骑手最后一行数据即可。

执行SQL

select
    rider_id,
    concat_ws(',',customer_id_list) as customer_id_list
from
(select rider_id,
       delivery_time,
       customer_id,
       collect_list(customer_id)over(partition by rider_id order by delivery_time) as customer_id_list,
       row_number() over (partition by rider_id order by delivery_time desc) as rn
from t_delivery_orders
    ) t
where rn = 1

执行结果

+-----------+---------------------------+
| rider_id  |     customer_id_list      |
+-----------+---------------------------+
| r001      | c001,c007,c023            |
| r002      | c003,c005,c002,c031,c001  |
| r003      | c004,c010                 |
+-----------+---------------------------+

2.拼接排序后拆分

2.1先把时间和顾客id进行拼接，拼接后进行行转列

执行SQL

select rider_id,
       delivery_time,
       customer_id,
       concat(delivery_time, customer_id) as time_customer
from t_delivery_orders

执行结果

+-----------+----------------------+--------------+--------------------------+
| rider_id  |    delivery_time     | customer_id  |      time_customer       |
+-----------+----------------------+--------------+--------------------------+
| r001      | 2024-01-01 08:30:00  | c001         | 2024-01-01 08:30:00c001  |
| r002      | 2024-01-01 08:43:00  | c003         | 2024-01-01 08:43:00c003  |
| r003      | 2024-01-01 09:15:00  | c004         | 2024-01-01 09:15:00c004  |
| r002      | 2024-01-01 09:21:00  | c005         | 2024-01-01 09:21:00c005  |
| r001      | 2024-01-01 09:30:00  | c007         | 2024-01-01 09:30:00c007  |
| r002      | 2024-01-01 09:47:00  | c002         | 2024-01-01 09:47:00c002  |
| r003      | 2024-01-01 10:21:00  | c010         | 2024-01-01 10:21:00c010  |
| r001      | 2024-01-01 11:56:00  | c023         | 2024-01-01 11:56:00c023  |
| r002      | 2024-01-01 12:30:00  | c031         | 2024-01-01 12:30:00c031  |
| r002      | 2024-01-01 13:30:00  | c001         | 2024-01-01 13:30:00c001  |
+-----------+----------------------+--------------+--------------------------+

2.2 对time_customer 按照骑手id分组行转列,并使用sort_array排序

执行SQL

select rider_id,
       sort_array(collect_list(time_customer)) as sorted_time_customer
from (select rider_id,
             delivery_time,
             customer_id,
             concat(delivery_time, customer_id) as time_customer
      from t_delivery_orders) t
group by rider_id

执行结果

+-----------+----------------------------------------------------+
| rider_id  |                sorted_time_customer                |
+-----------+----------------------------------------------------+
| r001      | ["2024-01-01 08:30:00c001","2024-01-01 09:30:00c007","2024-01-01 11:56:00c023"] |
| r002      | ["2024-01-01 08:43:00c003","2024-01-01 09:21:00c005","2024-01-01 09:47:00c002","2024-01-01 12:30:00c031","2024-01-01 13:30:00c001"] |
| r003      | ["2024-01-01 09:15:00c004","2024-01-01 10:21:00c010"] |
+-----------+----------------------------------------------------+

2.3 去掉时间部分后，转换成字符串

执行SQL

select rider_id,
       concat_ws(',', transform(sort_array(collect_list(time_customer)), x->substr(x, 20))) as customer_list
from (select rider_id,
             delivery_time,
             customer_id,
             concat(delivery_time, customer_id) as time_customer
      from t_delivery_orders) t
group by rider_id

执行结果

+-----------+---------------------------+
| rider_id  |       customer_list       |
+-----------+---------------------------+
| r001      | c001,c007,c023            |
| r002      | c003,c005,c002,c031,c001  |
| r003      | c004,c010                 |
+-----------+---------------------------+

四、数据准备

--建表语句
CREATE TABLE IF NOT EXISTS t_delivery_orders
(
    rider_id      string,         -- 骑手ID
    order_id      string,         -- 订单ID
    goods_type    STRING,         -- 物品类型
    delivery_time STRING,         -- 送达时间
    customer_id   STRING,         -- 客户id
    distance      decimal(10, 2), -- 配送距离，单位可以是公里或英里
    payment       decimal(10, 2)  -- 支付金额，骑手的配送费用
)
    COMMENT '骑手配送订单表';
--插入数据
INSERT INTO t_delivery_orders VALUES
('r001', '0001', '食品', '2024-01-01 08:30:00','c001',8.05,7.50),
('r002', '0002', '酒水', '2024-01-01 08:43:00','c003',3.01,13.00),
('r003', '0003', '食品', '2024-01-01 09:15:00','c004',4.12,3.50),
('r002', '0004', '文件', '2024-01-01 09:21:00','c005',10.98,15.00),
('r001', '0005', '食品', '2024-01-01 09:30:00','c007',2.32,5.00),
('r002', '0006', '食品', '2024-01-01 09:47:00','c002',0.78,5.00),
('r003', '0007', '文件', '2024-01-01 10:21:00','c010',8.11,8.00),
('r001', '0008', '酒水', '2024-01-01 11:56:00','c023',4.35,15.00),
('r002', '0009', '电子产品', '2024-01-01 12:30:00','c031',5.05,9.50),
('r002', '0010', '文件', '2024-01-01 13:30:00','c001',6.05,7.00);

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原始发表：2024-08-12，如有侵权请联系 cloudcommunity@tencent.com 删除

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