CODEWITHSUNDEEP
rpredictioninterceptglmnet
Peter S
Peter S

Reputation: 55

GLMNET prediction with intercept

I have two questions about prediction using GLMNET - specifically about the intercept.

I made a small example of train data creation, GLMNET estimation and prediction on the train data (which I will later change to Test data):

# Train data creation
Train <- data.frame('x1'=runif(10), 'x2'=runif(10))
Train$y <- Train$x1-Train$x2+runif(10)
# From Train data frame to x and y matrix
y <- Train$y
x <- as.matrix(Train[,c('x1','x2')])
# Glmnet model
Model_El <- glmnet(x,y)
Cv_El <- cv.glmnet(x,y)
# Prediction
Test_Matrix <- model.matrix(~.-y,data=Train)[,-1]
Test_Matrix_Df <- data.frame(Test_Matrix)
Pred_El <- predict(Model_El,newx=Test_Matrix,s=Cv_El$lambda.min,type='response')

I want to have an intercept in the estimated formula. This code gives an error concerning the dimensions of the Test_Matrix matrix unless I remove the (Intercept) column of the matrix - as in 

Test_Matrix <- model.matrix(~.-y,data=Train)[,-1]

My questions are:

Thanks in advance.

Upvotes: 1

Views: 4456

Answers (2)

jav
jav

Reputation: 1495

The matrix x you were feeding into the glmnet function doesn't contain an intercept column. Therefore, you should respect this format when constructing your test matrix: i.e. just do model.matrix(y ~ . - 1, data = Train).

By default, an intercept is fit in glmnet (see the intercept parameter in the glmnet function). Therefore, when you called glmnet(x, y), you are technically doing glmnet(x, y, intercept = T). Thus, even though your x matrix didn't have an intercept, one was fit for you.

Upvotes: 3

Zheyuan Li
Zheyuan Li

Reputation: 73325

If you want to predict a model with intercept, you have to fit a model with intercept. Your code used model matrix x <- as.matrix(Train[,c('x1','x2')]) which is intercept-free, therefore if you provide an intercept when using predict, you get an error.

You can do the following:

x <- model.matrix(y ~ ., Train)    ## model matrix with intercept
Model_El <- glmnet(x,y)
Cv_El <- cv.glmnet(x,y)
Test_Matrix <- model.matrix(y ~ ., Train)   ## prediction matrix with intercept
Pred_El <- predict(Model_El, newx = Test_Matrix, s = Cv_El$lambda.min, type='response')

Note, you don't have to do

model.matrix(~ . -y)

model.matrix will ignore the LHS of the formula, so it is legitimate to use

model.matrix(y ~ .)

Upvotes: 2

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