Teradata SQL：如果满足条件，则计算运行总计

Question

我有一个包含以下列和数据的数据集：

Customer | Week_number | Amount
cust1    |  0          | 100
cust1    |  1          | 200
cust1    |  3          | 300
cust2    |  0          | 1000
cust2    |  1          | 2000

我需要为每个客户计算每两周的总数。

使用窗口函数，我可以做到这一点：

SELECT 
 CUSTOMER, WEEK_NUMBER
, SUM(AMOUNT) OVER (PARTITION BY CUSTOMER ORDER BY WEEK_NUMBER ROWS 1 PRECEDING) AS FORTNIGHT_AMOUNT
FROM AMOUNT

但是，即使前一周没有金额，这也会增加金额。在上面的示例中，对于第 3 行的 cust1，它将第 3 周和第 1 周相加。只有当 week_number 比当前行的周数小 1 时，才应添加金额。这可能吗？感谢您的帮助。

我得到了什么：

Customer | Week_number | Fortnight_Amount
cust1    |  0          | 100
cust1    |  1          | 300
cust1    |  3          | **500**
cust2    |  0          | 1000
cust2    |  1          | 3000

预期结果：

Customer | Week_number | Fortnight_Amount
cust1    |  0          | 100
cust1    |  1          | 300
cust1    |  3          | **300**
cust2    |  0          | 1000
cust2    |  1          | 3000

Answer 1

如果您只想忽略不立即连续的周数，您可以先使用lag()，然后再创建一个窗口sum()：

select
    customer,
    week_number,
    sum(
        case when lag_week_number is null or week_number = lag_week_number + 1 
            then amount
            else 0 
        end
    ) over(partition by customer order by week_number) fortnight_amount
from (
    select 
        t.*, 
        lag(week_number) over(partition by customer order by week_number) lag_week_number
    from mytable t
) t

实际上，您可能真的想重置 sum，当 week_numbers 存在差距时。为此，这是某种间隙和孤岛分配，您将采用不同的方式进行操作：想法是当两个连续的周数按顺序排列时进行累积 sum 以开始一个新组，然后在每个组内求和：

select 
    customer,
    week_number,
    sum(amount) over(partition by customer, grp order by week_date) fortnight_amount
from (
    select 
        t.*,
        sum(
            case 
                when lag_week_number is null or week_number = lag_week_number + 1 
                then 0
                else 1
            end
        ) grp
    from (
        select 
            t.*, 
            lag(week_number) over(partition by customer order by week_number) lag_week_number
        from mytable t
    ) t
) t

Answer 2

您想要range 分区，而不是row 分区：

SELECT CUSTOMER, WEEK_NUMBER,
       SUM(AMOUNT) OVER (PARTITION BY CUSTOMER
                         ORDER BY WEEK_NUMBER 
                         RANGE BETWEEN 1 PRECEDING AND CURRENT ROW
                        ) AS FORTNIGHT_AMOUNT
FROM AMOUNT;

Answer 3

感谢@Gordon 和@GMB 的回答。不幸的是，我无法在 Teradata SQL 中同时使用 LAG 函数或 RANGE 分区。但我能够使用你们所描述的概念来获得以下答案。

SELECT 
CUSTOMER
, WEEK_NUMBER
, LAG_WEEK_NUMBER
, AMOUNT
, CASE 
  WHEN WEEK_NUMBER = LAG_WEEK_NUMBER + 1 
  THEN SUM(AMOUNT) OVER (PARTITION BY CUSTOMER ORDER BY WEEK_NUMBER ROWS BETWEEN 1 PRECEDING AND CURRENT ROW)
  ELSE AMOUNT
END AS TWO_WEEK_SUM_AMOUNT
FROM (
  SELECT 
  T.*
  , MAX(WEEK_NUMBER) OVER (PARTITION BY CUSTOMER ORDER BY WEEK_NUMBER ROWS BETWEEN 1 PRECEDING AND 1 PRECEDING) AS LAG_WEEK_NUMBER
  FROM MY_TABLE T
  ) T
ORDER BY CUSTOMER, WEEK_NUMBER

我能够从@dnoeth 在这些链接中的回答中获得 Teradata 中的 LAG 函数实现：

MAX(WEEK_NUMBER) OVER (PARTITION BY CUSTOMER ORDER BY WEEK_NUMBER ROWS BETWEEN 1 PRECEDING AND 1 PRECEDING) AS LAG_WEEK_NUMBER

rows between 1 preceding and preceding 1

Teradata partitioned query ... following rows dynamically

如果您发现答案有任何问题或是否可以通过任何方式改进，请告诉我。

Answer 4

如果只有两周/行，您的查询可以进一步简化为解释中的单个 STATS 步骤（因为两个 OLAP 函数都应用相同的 PARTITION/ORDER）：

SELECT T.*
, CASE 
    WHEN MAX(WEEK_NUMBER) OVER (PARTITION BY CUSTOMER ORDER BY WEEK_NUMBER ROWS BETWEEN 1 PRECEDING AND 1 PRECEDING) + 1 = WEEK_NUMBER
    THEN SUM(AMOUNT)      OVER (PARTITION BY CUSTOMER ORDER BY WEEK_NUMBER ROWS BETWEEN 1 PRECEDING AND CURRENT ROW)
   ELSE AMOUNT
  END AS TWO_WEEK_SUM_AMOUNT
FROM MY_TABLE T
ORDER BY CUSTOMER, WEEK_NUMBER

当然，这假设周从 0 开始，并且没有上一年的第 52/53 周。