我在sql数据库中有一些数据，我想计算斜率。数据具有以下布局：
Date | Keyword | Score 2012-01-10 | ipad | 0.12 2012-01-11 | ipad | 0.17 2012-01-12 | ipad | 0.24 2012-01-10 | taco | 0.19 2012-01-11 | taco | 0.34 2012-01-12 | taco | 0.45
我希望通过使用SQL创建新表来使最终输出看起来像这样：
Date | Keyword | Score | Slope 2012-01-10 | ipad | 0.12 | 0.06 2012-01-11 | ipad | 0.17 | 0.06 2012-01-12 | ipad | 0.24 | 0.06 2012-01-10 | taco | 0.19 | 0.13 2012-01-11 | taco | 0.34 | 0.13 2012-01-12 | taco | 0.45 | 0.13
为了使事情复杂化，并非所有的关键字都具有3个日期的数据，例如，有些只有2个日期。
SQL越简单越好，因为我的数据库是专有的，并且我不确定是否可以使用哪些公式，尽管我知道它可以执行OVER（PARTITION BY）。谢谢！
更新：我将斜率定义为最佳拟合y = mx + p，在excel中将是= slope（）
这是我通常在excel中操作的另一个实际示例：
date keyword score slope 1/22/2012 water bottle 0.010885442 0.000334784 1/23/2012 water bottle 0.011203949 0.000334784 1/24/2012 water bottle 0.008460835 0.000334784 1/25/2012 water bottle 0.010363991 0.000334784 1/26/2012 water bottle 0.011800716 0.000334784 1/27/2012 water bottle 0.012948411 0.000334784 1/28/2012 water bottle 0.012732459 0.000334784 1/29/2012 water bottle 0.011682568 0.000334784
我能做的最干净的一个：
SELECT Scores.Date, Scores.Keyword, Scores.Score, (N * Sum_XY - Sum_X * Sum_Y)/(N * Sum_X2 - Sum_X * Sum_X) AS Slope FROM Scores INNER JOIN ( SELECT Keyword, COUNT(*) AS N, SUM(CAST(Date as float)) AS Sum_X, SUM(CAST(Date as float) * CAST(Date as float)) AS Sum_X2, SUM(Score) AS Sum_Y, SUM(Score*Score) AS Sum_Y2, SUM(CAST(Date as float) * Score) AS Sum_XY FROM Scores GROUP BY Keyword ) G ON G.Keyword = Scores.Keyword;
它使用简单线性回归来计算斜率。
结果：
Date Keyword Score Slope 2012-01-22 water bottle 0,010885442 0,000334784345222076 2012-01-23 water bottle 0,011203949 0,000334784345222076 2012-01-24 water bottle 0,008460835 0,000334784345222076 2012-01-25 water bottle 0,010363991 0,000334784345222076 2012-01-26 water bottle 0,011800716 0,000334784345222076 2012-01-27 water bottle 0,012948411 0,000334784345222076 2012-01-28 water bottle 0,012732459 0,000334784345222076 2012-01-29 water bottle 0,011682568 0,000334784345222076
每个数据库系统似乎都有不同的方法将日期转换为数字：
TO_SECONDS(date)
TO_DAYS(date)
TO_NUMBER(TO_CHAR(date, 'J'))
date - TO_DATE('1','yyyy')
CAST(date AS float)
CONVERT
