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MASS :: lm.ridge系数

[英]MASS::lm.ridge coefficients

I'm trying to perform ridge regression using the lm.ridge method. 我正在尝试使用lm.ridge方法执行岭回归。 My question is how can I get the coefficients of the fitted model? 我的问题是如何获得拟合模型的系数? I'm getting different results from calling: 我从致电获得不同的结果:

  • model$coef
  • coef(model)

Which is the correct one? 哪个是正确的? Also, why do I get different results from calling: 另外,为什么我从调用中得到不同的结果:

  • coef(model) and looking at the 1st coefficient, vs. coef(model)coef(model)系数vs.
  • coef(model)[1] ? coef(model)[1]

As ?lm.ridge says (in describing the $coef element of the returned object) [emphasis added] 正如?lm.ridge所说(在描述返回对象的$coef元素时)[强调加]

coef: matrix of coefficients, one row for each value of 'lambda'. coef:系数矩阵,每个“ lambda”值对应一行。 Note that these are not on the original scale and are for use by the 'coef' method. 请注意,这些不是原始比例,而是由“ coef”方法使用。

This means, specifically, that the $coef element is not intended for end-users ("if you have to ask ..."). 具体地说,这意味着$coef元素不适合最终用户使用(“如果您要询问...”)。 (If you want to see how $coef is translated, inspect MASS:::coef.ridgelm .) In general, it's better practice to use an accessor method such as coef() , when it exists, than to extract components from the guts of a returned object using $ (or @ for S4 objects) - for exactly this reason. (如果您想了解$coef的翻译方式,请检查MASS:::coef.ridgelm 。)通常,最好使用访问器方法(例如coef() (如果存在),而不是从胆量中提取成分。正是由于这个原因,使用$ (或对于S4对象为@对返回的对象进行计数。 Package authors provide coef() methods for a reason ... 程序包作者出于某种原因提供了coef()方法。

I can't replicate your second question. 我不能重复你的第二个问题。 Using the model from ?lm.ridge , the answers seem identical except for the precision with which they are printed ... 使用来自?lm.ridge的模型,答案似乎相同,除了它们的打印精度...


> m1 <- lm.ridge(y ~ ., longley)
> coef(m1)
                            GNP    Unemployed  Armed.Forces    Population 
    2946.85636017    0.26352725    0.03648291    0.01116105   -1.73702984 
             Year      Employed 
      -1.41879853    0.23128785 
> coef(m1)[1]       
2946.856

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