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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