我应该如何获得套索模型的系数？

Question

这是我的代码：

library(MASS)
library(caret)
df <- Boston
set.seed(3721)
cv.10.folds <- createFolds(df$medv, k = 10)
lasso_grid <- expand.grid(fraction=c(1,0.1,0.01,0.001))
lasso <- train(medv ~ ., 
               data = df, 
               preProcess = c("center", "scale"),
               method ='lasso',
               tuneGrid = lasso_grid,
               trControl= trainControl(method = "cv", 
                                       number = 10, 
                                       index = cv.10.folds))  

lasso

与线性模型不同，我无法从摘要（套索）中找到套索回归模型的系数。我该怎么做？或者我可以使用 glmnet 吗？

Answer 1

当您使用 进行训练时method="lasso"，elasticnet 中的 enet 被调用：

lasso$finalModel$call
elasticnet::enet(x = as.matrix(x), y = y, lambda = 0)

小插图写道：

LARS-EN 算法以与最小二乘拟合相同的计算成本同时计算收缩参数的所有值的完整弹性网络解

在 下lasso$finalModel$beta.pure，您拥有与 下 L1 范数的 16 个值相对应的所有 16 组系数的系数lasso$finalModel$L1norm：

length(lasso$finalModel$L1norm)
[1] 16

dim(lasso$finalModel$beta.pure)
[1] 16 13

您也可以使用预测来查看它：

predict(lasso$finalModel,type="coef")
$s
 [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16

$fraction
 [1] 0.00000000 0.06666667 0.13333333 0.20000000 0.26666667 0.33333333
 [7] 0.40000000 0.46666667 0.53333333 0.60000000 0.66666667 0.73333333
[13] 0.80000000 0.86666667 0.93333333 1.00000000

$mode
[1] "step"

$coefficients
          crim        zn       indus      chas        nox       rm        age
0   0.00000000 0.0000000  0.00000000 0.0000000  0.0000000 0.000000 0.00000000
1   0.00000000 0.0000000  0.00000000 0.0000000  0.0000000 0.000000 0.00000000
2   0.00000000 0.0000000  0.00000000 0.0000000  0.0000000 1.677765 0.00000000
3   0.00000000 0.0000000  0.00000000 0.0000000  0.0000000 2.571071 0.00000000
4   0.00000000 0.0000000  0.00000000 0.0000000  0.0000000 2.716138 0.00000000
5   0.00000000 0.0000000  0.00000000 0.2586083  0.0000000 2.885615 0.00000000
6  -0.05232643 0.0000000  0.00000000 0.3543411  0.0000000 2.953605 0.00000000
7  -0.13286554 0.0000000  0.00000000 0.4095229  0.0000000 2.984026 0.00000000
8  -0.21665925 0.0000000  0.00000000 0.5196189 -0.5933941 3.003512 0.00000000
9  -0.32168140 0.3326103  0.00000000 0.6044308 -1.0246080 2.973693 0.00000000
10 -0.33568474 0.3771889 -0.02165730 0.6165190 -1.0728128 2.967696 0.00000000
11 -0.42820289 0.4522827 -0.09212253 0.6407298 -1.2474934 2.932427 0.00000000
12 -0.62605363 0.7005114  0.00000000 0.6574277 -1.5655601 2.832726 0.00000000
13 -0.88747102 1.0150162  0.00000000 0.6856705 -1.9476465 2.694820 0.00000000
14 -0.91679342 1.0613165  0.09956489 0.6837833 -2.0217269 2.684401 0.00000000
15 -0.92906457 1.0826390  0.14103943 0.6824144 -2.0587536 2.676877 0.01948534

由插入符调整的超参数是最大 L1 范数的分数，因此在您提供的结果中，它将是 1，即 max ：

lasso
The lasso 

506 samples
 13 predictor

Pre-processing: centered (13), scaled (13) 
Resampling: Cross-Validated (10 fold) 
Summary of sample sizes: 51, 51, 51, 50, 51, 50, ... 
Resampling results across tuning parameters:

  fraction  RMSE      Rsquared   MAE     
  0.001     9.182599  0.5075081  6.646013
  0.010     9.022117  0.5075081  6.520153
  0.100     7.597607  0.5572499  5.402851
  1.000     6.158513  0.6033310  4.140362

RMSE was used to select the optimal model using the smallest value.
The final value used for the model was fraction = 1.

要获得最佳分数的系数：

predict(lasso$finalModel,type="coef",s=16)
$s
[1] 16

$fraction
[1] 1

$mode
[1] "step"

$coefficients
       crim          zn       indus        chas         nox          rm 
-0.92906457  1.08263896  0.14103943  0.68241438 -2.05875361  2.67687661 
        age         dis         rad         tax     ptratio       black 
 0.01948534 -3.10711605  2.66485220 -2.07883689 -2.06264585  0.85010886 
      lstat 
-3.74733185