在“最佳" lambda 处获取 glmnet 系数 [英] Getting glmnet coefficients at 'best' lambda

问题描述

我在 glmnet 中使用以下代码:

I am using following code with glmnet:

> library(glmnet)
> fit = glmnet(as.matrix(mtcars[-1]), mtcars[,1])
> plot(fit, xvar='lambda')

但是，我想打印出最好的 Lambda 系数，就像在岭回归中所做的那样.我看到以下拟合结构:

However, I want to print out the coefficients at best Lambda, like it is done in ridge regression. I see following structure of fit:

> str(fit)
List of 12
 $ a0       : Named num [1:79] 20.1 21.6 23.2 24.7 26 ...
  ..- attr(*, "names")= chr [1:79] "s0" "s1" "s2" "s3" ...
 $ beta     :Formal class 'dgCMatrix' [package "Matrix"] with 6 slots
  .. ..@ i       : int [1:561] 0 4 0 4 0 4 0 4 0 4 ...
  .. ..@ p       : int [1:80] 0 0 2 4 6 8 10 12 14 16 ...
  .. ..@ Dim     : int [1:2] 10 79
  .. ..@ Dimnames:List of 2
  .. .. ..$ : chr [1:10] "cyl" "disp" "hp" "drat" ...
  .. .. ..$ : chr [1:79] "s0" "s1" "s2" "s3" ...
  .. ..@ x       : num [1:561] -0.0119 -0.4578 -0.1448 -0.7006 -0.2659 ...
  .. ..@ factors : list()
 $ df       : int [1:79] 0 2 2 2 2 2 2 2 2 3 ...
 $ dim      : int [1:2] 10 79
 $ lambda   : num [1:79] 5.15 4.69 4.27 3.89 3.55 ...
 $ dev.ratio: num [1:79] 0 0.129 0.248 0.347 0.429 ...
 $ nulldev  : num 1126
 $ npasses  : int 1226
 $ jerr     : int 0
 $ offset   : logi FALSE
 $ call     : language glmnet(x = as.matrix(mtcars[-1]), y = mtcars[, 1])
 $ nobs     : int 32
 - attr(*, "class")= chr [1:2] "elnet" "glmnet"

但我无法获得最佳的 Lambda 和相应的系数.感谢您的帮助.

But I am not able to get the best Lambda and the corresponding coefficients. Thanks for your help.

推荐答案

试试这个:

fit = glmnet(as.matrix(mtcars[-1]), mtcars[,1], 
    lambda=cv.glmnet(as.matrix(mtcars[-1]), mtcars[,1])$lambda.1se)
coef(fit)

或者你可以在coef中指定一个指定的lambda值:

Or you can specify a specify a lambda value in coef:

fit = glmnet(as.matrix(mtcars[-1]), mtcars[,1])
coef(fit, s = cv.glmnet(as.matrix(mtcars[-1]), mtcars[,1])$lambda.1se)

您需要选择一个最佳" lambda，而 lambda.1se 是一个合理的选择.但是您可以使用 cv.glmnet(as.matrix(mtcars[-1]), mtcars[,1])$lambda.min 或您认为最佳"的任何其他 lambda 值给你.

You need to pick a "best" lambda, and lambda.1se is a reasonable, or justifiable, one to pick. But you could use cv.glmnet(as.matrix(mtcars[-1]), mtcars[,1])$lambda.min or any other value of lambda that you settle upon as "best" for you.

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