sql中的指数衰减

2023年1月25日 198次阅读

请考虑下表：

Person | 1/1/13 | 1/2/13  | 1/3/13 | 1/4/13 | 1/5/13
Bill   | 4      | 2       | 1      | .5     | .25
Jane   | 0      | 0       | 2      | 1      | .5
Mary   | 0      | 8       | 4      | 2      | 1
-------------------------------------------------
Total  | 4      | 10      | 7      | 3.5    | 1.75

这来自下表：

Bill | 1/1/13 | 4
Jane | 1/3/13 | 2
Mary | 1/2/13 | 8

基本上,我们知道第一天,然后我们假设每隔一天减少一半的价值.我想从第一个表中获取“总”行.

有没有办法在(T-)SQL中执行此操作？我已经在R中创建了它,但我完全不知道如何在SQL中实现它. (日期是实际日期,而不仅仅是星期几.)

最佳答案 如果你的表看起来像这样：

CREATE TABLE t (person VARCHAR(7), day_of_week_name VARCHAR(7), value NUMERIC);
INSERT INTO t VALUES ('Bill', 'Monday', 4);
INSERT INTO t values ('Jane', 'Weds', 2);
INSERT INTO t VALUES ('Mary', 'Tuesday', 8);

你有一些day_of_week表与天的相对位置：

CREATE TABLE day_of_week (name VARCHAR(7), position INT);
INSERT INTO day_of_week VALUES ('Monday', 1);
INSERT INTO day_of_week values ('Tuesday', 2);
INSERT INTO day_of_week VALUES ('Weds', 3);
INSERT INTO day_of_week VALUES ('Thurs', 4);
INSERT INTO day_of_week VALUES ('Friday', 5);

然后用PIVOT做这个并不太难看：

SELECT Monday, Tuesday, Weds, Thurs, Friday
  FROM ( SELECT dow2.name AS day_of_week_name,
                t.value / power(2, dow2.position - dow1.position) AS decayed_value
           FROM t
           JOIN day_of_week AS dow1
             ON t.day_of_week_name = dow1.name
           JOIN day_of_week AS dow2
             ON dow1.position <= dow2.position
       ) AS b
 PIVOT ( SUM(decayed_value)
           FOR day_of_week_name
             IN (Monday, Tuesday, Weds, Thurs, Friday)
       ) AS pvt
;

(SQL Fiddle here.)

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