dat*_*oda 14 mysql sql oracle10g
我在sql数据库中有一些数据,我想计算斜率.数据有这样的布局:
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
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我想通过使用SQL创建一个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
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更复杂的是,并非所有关键字都有3个日期值的数据,有些只有2个例如.
SQL越简单越好,因为我的数据库是专有的,我不太确定哪些公式可用,虽然我知道如果有帮助我可以做OVER(PARTITION BY).谢谢!
更新:我将斜率定义为最佳拟合y = mx + p又名在excel中它将是= 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
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Mar*_*rot 15
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;
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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
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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)或TO_DAYS(date)TO_NUMBER(TO_CHAR(date, 'J'))或date - TO_DATE('1','yyyy')CAST(date AS float)或等效的CONVERT)| 归档时间: |
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