Teradata SQL：如果滿足條件，則計算運行總數

Question

我有一個包含以下列和數據的數據集：

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

我需要為每個客戶每兩周計算一次總數。

使用窗口函數，我能夠做到這一點：

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

但即使前一周沒有金額，這也會增加金額。 在上面的示例中，對於 cust1，第 3 行，它將第 3 周和第 1 周相加。僅當 week_number 比當前行的周少 1 時才應添加數量。 這可能嗎？ 謝謝您的幫助。

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

預期結果：

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

Answer 1

如果只有兩周/行，您的查詢可以進一步簡化為解釋中的單個 STATS 步驟（因為兩個 OLAP 函數都應用相同的 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

當然，這假設周從 0 開始，並且沒有前一年的第 52/53 周。

Answer 2

如果您只想忽略不是立即順序的周數，您可以先使用lag() ，然后執行窗口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

實際上，當周數存在差距時，您實際上可能想要重置sum 。 為此，這是某種間隙和孤島分配，您將采取不同的方式：這個想法是在兩個連續的周數 ae 順序時進行累積sum以開始一個新組，然后在每個組內求和：

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 3

您需要range分區，而不是row分區：

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 4

感謝@Gordon 和@GMB 的回答。 不幸的是，我無法在 Teradata SQL 中同時使用 LAG 函數或 RANGE 分區。 但是我能夠使用你們描述的概念來獲得以下答案。

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

我能夠從這些鏈接中@dnoeth 的回答中獲得 Teradata 中的 LAG 函數實現：

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

前 1 行和前 1 行之間的行

Teradata 分區查詢...動態跟蹤行

如果您發現答案有任何問題或者是否可以以任何方式改進，請告訴我。