Teradata SQL：如果满足条件则计算运行总数

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

我需要计算每个客户每两周的总计。

使用 window 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，这也会加起来。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。仅当 week_number 比当前行的周数小 1 时才应添加该金额。这可能吗？感谢您的帮助。

我得到的是：

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

如果您只想忽略不是立即连续的周数，您可以先使用 lag()，然后再执行 window 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，当 week_numbers 中存在间隙时，您可能实际上想要重置 sum。对于这个，这是某种差距和岛屿分配，你会以不同的方式进行：这个想法是做一个累积 sum 来开始一个新的组，当两个连续的周数是连续的，然后求和每组：

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。但我能够使用你们描述的概念得到以下答案。

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可以通过任何方式改进，请告诉我。

Answer 4

如果只有两个 weeks/rows，您的查询可以在 Explain 中进一步简化为单个统计步骤（因为两个 OLAP 函数应用相同 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。

Teradata SQL：如果满足条件则计算运行总数

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Teradata SQL：如果满足条件则计算 运行 总数

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Teradata SQL：如果满足条件则计算运行总数