mysql - 最优二元线性回归计算

Question

问题

我希望将 y = mx + b 方程(其中 m 是 SLOPE，b 是 INTERCEPT)应用于数据集，如 SQL 代码所示检索。 (MySQL) 查询的值是:

SLOPE = 0.0276653965651912
INTERCEPT = -57.2338357550468

SQL 代码

SELECT
  ((sum(t.YEAR) * sum(t.AMOUNT)) - (count(1) * sum(t.YEAR * t.AMOUNT))) /
  (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as SLOPE,

  ((sum( t.YEAR ) * sum( t.YEAR * t.AMOUNT )) -
  (sum( t.AMOUNT ) * sum(power(t.YEAR, 2)))) /
  (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as INTERCEPT,
FROM
(SELECT
  D.AMOUNT,
  Y.YEAR
FROM
  CITY C, STATION S, YEAR_REF Y, MONTH_REF M, DAILY D
WHERE
  -- For a specific city ...
  --
  C.ID = 8590 AND
  -- Find all the stations within a 15 unit radius ...
  --
  SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) < 15 AND
  -- Gather all known years for that station ...
  --
  S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND
  -- The data before 1900 is shaky; insufficient after 2009.
  --
  Y.YEAR BETWEEN 1900 AND 2009 AND
  -- Filtered by all known months ...
  --
  M.YEAR_REF_ID = Y.ID AND
  -- Whittled down by category ...
  --
  M.CATEGORY_ID = '001' AND
  -- Into the valid daily climate data.
  --
  M.ID = D.MONTH_REF_ID AND
  D.DAILY_FLAG_ID <> 'M'
  GROUP BY Y.YEAR
  ORDER BY Y.YEAR
) t

问题

以下结果(计算线的起点和终点)显示不正确。为什么结果偏离了约 10 度(例如，异常值扭曲了数据)？

(1900 * 0.0276653965651912) + (-57.2338357550468) = -4.66958228

(2009 * 0.0276653965651912) + (-57.2338357550468) = -1.65405406

(请注意，数据不再与图像匹配；代码。)

我原本预计 1900 年的结果大约是 10(不是 -4.67)，而 2009 年的结果是大约 11.50(不是 -1.65)。

相关网站

• Least absolute deviations
• Robust regression

Answer 1

尝试拆分函数，你计算错了参数。看看here供引用。

我会做类似下面的事情(请原谅我不太记得 SQL 语法和临时变量，所以代码实际上可能是错误的):

SELECT

sum(t.YEAR) / count(1) AS avgX,

sum(t.AMOUNT) / count(1) AS avgY,

sum(t.AMOUNT*t.YEAR) / count(1) AS avgXY,

sum(power(t.YEAR, 2)) / count(1) AS avgXsq,

( avgXY - avgX * avgY ) / ( avgXsq - power(avgX, 2) )  as SLOPE,

avgY - SLOPE * avgX as INTERCEPT,