本文会从一个商务分析案例入手，说明SQL窗口函数的使用方式。通过本文的5个需求分析，可以看出SQL窗口函数的功能十分强大，不仅能够使我们编写的SQL逻辑更加清晰，而且在某种程度上可以简化需求开发。
数据准备
本文主要分析只涉及一张订单表orders，操作过程在Hive中完成，具体数据如下：
-- 建表
CREATE TABLE orders(
order_id int,
customer_id string,
city string,
add_time string,
amount decimal(10,2));
-- 准备数据
INSERT INTO orders VALUES
(1,"A","上海","2020-01-01 00:00:00.000000",200),
(2,"B","上海","2020-01-05 00:00:00.000000",250),
(3,"C","北京","2020-01-12 00:00:00.000000",200),
(4,"A","上海","2020-02-04 00:00:00.000000",400),
(5,"D","上海","2020-02-05 00:00:00.000000",250),
(5,"D","上海","2020-02-05 12:00:00.000000",300),
(6,"C","北京","2020-02-19 00:00:00.000000",300),
(7,"A","上海","2020-03-01 00:00:00.000000",150),
(8,"E","北京","2020-03-05 00:00:00.000000",500),
(9,"F","上海","2020-03-09 00:00:00.000000",250),
(10,"B","上海","2020-03-21 00:00:00.000000",600);需求1：收入增长
在业务方面，第m1个月的收入增长计算如下：100 *（m1-m0）/ m0
其中，m1是给定月份的收入，m0是上个月的收入。因此，从技术上讲，我们需要找到每个月的收入，然后以某种方式将每个月的收入与上一个收入相关联，以便进行上述计算。计算当时如下：
WITH
monthly_revenue as (
SELECT
trunc(add_time,'MM') as month,
sum(amount) as revenue
FROM orders
GROUP BY 1
)
,prev_month_revenue as (
SELECT
month,
revenue,
lag(revenue) over (order by month) as prev_month_revenue -- 上一月收入
FROM monthly_revenue
)
SELECT
month,
revenue,
prev_month_revenue,
round(100.0*(revenue-prev_month_revenue)/prev_month_revenue,1) as revenue_growth
FROM prev_month_revenue
ORDER BY 1结果输出
| month | revenue | prev_month_revenue | revenue_growth |
|---|---|---|---|
| 2020-01-01 | 650 | NULL | NULL |
| 2020-02-01 | 1250 | 650 | 92.3 |
| 2020-03-01 | 1500 | 1250 | 20 |
我们还可以按照按城市分组进行统计，查看某个城市某个月份的收入增长情况
WITH
monthly_revenue as (
SELECT
trunc(add_time,'MM') as month,
city,
sum(amount) as revenue
FROM orders
GROUP BY 1,2
)
,prev_month_revenue as (
SELECT
month,
city,
revenue,
lag(revenue) over (partition by city order by month) as prev_month_revenue
FROM monthly_revenue
)
SELECT
month,
city,
revenue,
round(100.0*(revenue-prev_month_revenue)/prev_month_revenue,1) as revenue_growth
FROM prev_month_revenue
ORDER BY 2,1结果输出
| month | city | revenue | revenue_growth |
|---|---|---|---|
| 2020-01-01 | 上海 | 450 | NULL |
| 2020-02-01 | 上海 | 950 | 111.1 |
| 2020-03-01 | 上海 | 1000 | 5.3 |
| 2020-01-01 | 北京 | 200 | NULL |
| 2020-02-01 | 北京 | 300 | 50 |
| 2020-03-01 | 北京 | 500 | 66.7 |
需求2：累计求和
累计汇总，即当前元素和所有先前元素的总和，如下面的SQL：
WITH
monthly_revenue as (
SELECT
trunc(add_time,'MM') as month,
sum(amount) as revenue
FROM orders
GROUP BY 1
)
SELECT
month,
revenue,
sum(revenue) over (order by month rows between unbounded preceding and current row) as running_total
FROM monthly_revenue
ORDER BY 1结果输出
| month | revenue | running_total |
|---|---|---|
| 2020-01-01 | 650 | 650 |
| 2020-02-01 | 1250 | 1900 |
| 2020-03-01 | 1500 | 3400 |
我们还可以使用下面的组合方式进行分析，SQL如下：
SELECT
order_id,
customer_id,
city,
add_time,
amount,
sum(amount) over () as amount_total, -- 所有数据求和
sum(amount) over (order by order_id rows between unbounded preceding and current row) as running_sum, -- 累计求和
sum(amount) over (partition by customer_id order by add_time rows between unbounded preceding and current row) as running_sum_by_customer,
avg(amount) over (order by add_time rows between 5 preceding and current row) as trailing_avg -- 滚动求平均
FROM orders
ORDER BY 1结果输出：
| order_id | customer_id | city | add_time | amount | amount_total | running_sum | running_sum_by_customer | trailing_avg |
|---|---|---|---|---|---|---|---|---|
| 1 | A | 上海 | 2020-01-01 00:00:00.000000 | 200 | 3400 | 200 | 200 | 200 |
| 2 | B | 上海 | 2020-01-05 00:00:00.000000 | 250 | 3400 | 450 | 250 | 225 |
| 3 | C | 北京 | 2020-01-12 00:00:00.000000 | 200 | 3400 | 650 | 200 | 216.666667 |
| 4 | A | 上海 | 2020-02-04 00:00:00.000000 | 400 | 3400 | 1050 | 600 | 262.5 |
| 5 | D | 上海 | 2020-02-05 00:00:00.000000 | 250 | 3400 | 1300 | 250 | 260 |
| 5 | D | 上海 | 2020-02-05 12:00:00.000000 | 300 | 3400 | 1600 | 550 | 266.666667 |
| 6 | C | 北京 | 2020-02-19 00:00:00.000000 | 300 | 3400 | 1900 | 500 | 283.333333 |
| 7 | A | 上海 | 2020-03-01 00:00:00.000000 | 150 | 3400 | 2050 | 750 | 266.666667 |
| 8 | E | 北京 | 2020-03-05 00:00:00.000000 | 500 | 3400 | 2550 | 500 | 316.666667 |
| 9 | F | 上海 | 2020-03-09 00:00:00.000000 | 250 | 3400 | 2800 | 250 | 291.666667 |
| 10 | B | 上海 | 2020-03-21 00:00:00.000000 | 600 | 3400 | 3400 | 850 |
需求3：处理重复数据
从上面的数据可以看出，存在两条重复的数据(5,"D","上海","2020-02-05 00:00:00.000000",250),
(5,"D","上海","2020-02-05 12:00:00.000000",300),显然需要对其进行清洗去重，保留最新的一条数据，SQL如下：
我们先进行分组排名，然后保留最新的那条数据即可：
SELECT *
FROM (
SELECT *,
row_number() over (partition by order_id order by add_time desc) as rank
FROM orders
) t
WHERE rank=1结果输出：
| t.order_id | t.customer_id | t.city | t.add_time | t.amount | t.rank |
|---|---|---|---|---|---|
| 1 | A | 上海 | 2020-01-01 00:00:00.000000 | 200 | 1 |
| 2 | B | 上海 | 2020-01-05 00:00:00.000000 | 250 | 1 |
| 3 | C | 北京 | 2020-01-12 00:00:00.000000 | 200 | 1 |
| 4 | A | 上海 | 2020-02-04 00:00:00.000000 | 400 | 1 |
| 5 | D | 上海 | 2020-02-05 12:00:00.000000 | 300 | 1 |
| 6 | C | 北京 | 2020-02-19 00:00:00.000000 | 300 | 1 |
| 7 | A | 上海 | 2020-03-01 00:00:00.000000 | 150 | 1 |
| 8 | E | 北京 | 2020-03-05 00:00:00.000000 | 500 | 1 |
| 9 | F | 上海 | 2020-03-09 00:00:00.000000 | 250 | 1 |
| 10 | B | 上海 | 2020-03-21 00:00:00.000000 | 600 | 1 |
经过上面的清洗过程，对数据进行了去重。重新计算上面的需求1，正确SQL脚本为：
WITH
orders_cleaned as (
SELECT *
FROM (
SELECT *,
row_number() over (partition by order_id order by add_time desc) as rank
FROM orders
)t
WHERE rank=1
)
,monthly_revenue as (
SELECT
trunc(add_time,'MM') as month,
sum(amount) as revenue
FROM orders_cleaned
GROUP BY 1
)
,prev_month_revenue as (
SELECT
month,
revenue,
lag(revenue) over (order by month) as prev_month_revenue
FROM monthly_revenue
)
SELECT
month,
revenue,
round(100.0*(revenue-prev_month_revenue)/prev_month_revenue,1) as revenue_growth
FROM prev_month_revenue
ORDER BY 1结果输出：
| month | revenue | revenue_growth |
|---|---|---|
| 2020-01-01 | 650 | NULL |
| 2020-02-01 | 1000 | 53.8 |
| 2020-03-01 | 1500 | 50 |
将清洗后的数据创建成视图，方便以后使用
CREATE VIEW orders_cleaned AS
SELECT
order_id,
customer_id,
city,
add_time,
amount
FROM (
SELECT *,
row_number() over (partition by order_id order by add_time desc) as rank
FROM orders
)t
WHERE rank=1需求4：分组取TopN
分组取topN是最长见的SQL窗口函数使用场景，下面的SQL是计算每个月份的top2订单金额，如下：
WITH orders_ranked as (
SELECT
trunc(add_time,'MM') as month,
*,
row_number() over (partition by trunc(add_time,'MM') order by amount desc, add_time) as rank
FROM orders_cleaned
)
SELECT
month,
order_id,
customer_id,
city,
add_time,
amount
FROM orders_ranked
WHERE rank <=2
ORDER BY 1需求5：重复购买行为
下面的SQL计算重复购买率：重复购买的人数/总人数*100%以及第一笔订单金额与第二笔订单金额之间的典型差额:avg(第二笔订单金额/第一笔订单金额)
WITH customer_orders as (
SELECT *,
row_number() over (partition by customer_id order by add_time) as customer_order_n,
lag(amount) over (partition by customer_id order by add_time) as prev_order_amount
FROM orders_cleaned
)
SELECT
round(100.0*sum(case when customer_order_n=2 then 1 end)/count(distinct customer_id),1) as repeat_purchases,-- 重复购买率
avg(case when customer_order_n=2 then 1.0*amount/prev_order_amount end) as revenue_expansion -- 重复购买较上次购买差异，第一笔订单金额与第二笔订单金额之间的典型差额
FROM customer_orders结果输出：
WITH结果输出：
| orders_cleaned.order_id | orders_cleaned.customer_id | orders_cleaned.city | orders_cleaned.add_time | orders_cleaned.amount | customer_order_n | prev_order_amount |
|---|---|---|---|---|---|---|
| 1 | A | 上海 | 2020-01-01 00:00:00.000000 | 200 | 1 | NULL |
| 4 | A | 上海 | 2020-02-04 00:00:00.000000 | 400 | 2 | 200 |
| 7 | A | 上海 | 2020-03-01 00:00:00.000000 | 150 | 3 | 400 |
| 2 | B | 上海 | 2020-01-05 00:00:00.000000 | 250 | 1 | NULL |
| 10 | B | 上海 | 2020-03-21 00:00:00.000000 | 600 | 2 | 250 |
| 3 | C | 北京 | 2020-01-12 00:00:00.000000 | 200 | 1 | NULL |
| 6 | C | 北京 | 2020-02-19 00:00:00.000000 | 300 | 2 | 200 |
| 5 | D | 上海 | 2020-02-05 12:00:00.000000 | 300 | 1 | NULL |
| 8 | E | 北京 | 2020-03-05 00:00:00.000000 | 500 | 1 | NULL |
| 9 | F | 上海 | 2020-03-09 00:00:00.000000 | 250 |
最终结果输出：
| repeat_purchases | revenue_expansion |
|---|---|
| 50 | 1.9666666666666668 |
总结
本文主要分享了SQL窗口函数的基本使用方式以及使用场景，并结合了具体的分析案例。通过本文的分析案例，可以加深对SQL窗口函数的理解。
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