Extracting coefficient variable names from glmnet into a data.frame

Question

I would like to extract the glmnet generated model coefficients and create a SQL query from them. The function coef(cv.glmnet.fit) yields a 'dgCMatrix' object. When I convert it to a matrix using as.matrix, the variable names are lost and only the coefficient values are left behind.

I know one can print the coefficients in the screen, however is it possible to write the names to a data frame?

Can anybody assist to extract these names?

Answer 1

UPDATE: Both first two comments of my answer are right. I have kept the answer below the line just for posterity.

The following answer is short, it works and does not need any other package:

tmp_coeffs <- coef(cv.glmnet.fit, s = "lambda.min")
data.frame(name = tmp_coeffs@Dimnames[[1]][tmp_coeffs@i + 1], coefficient = tmp_coeffs@x)

The reason for +1 is that the @i method indexes from 0 for the intercept but @Dimnames[[1]] starts at 1.

---

OLD ANSWER: (only kept for posterity) Try these lines:

The non zero coefficients:

coef(cv.glmnet.fit, s = "lambda.min")[which(coef(cv.glmnet.fit, s = "lambda.min") != 0)]

The features that are selected:

colnames(regression_data)[which(coef(cv.glmnet.fit, s = "lambda.min") != 0)]

Then putting them together as a dataframe is staight forward, but let me know if you want that part of the code also.

---

Answer 2

Check broom package. It has tidy function that converts output of different R objects (including glmnet) into data.frames.

Answer 3

The names should be accessible as dimnames(coef(cv.glmnet.fit))[[1]], so the following should put both coefficient names and values into a data.frame: data.frame(coef.name = dimnames(coef(GLMNET))[[1]], coef.value = matrix(coef(GLMNET)))

Answer 4

Building on Mehrad's solution above, here is a simple function to print a table containing only the non-zero coefficients:

print_glmnet_coefs <- function(cvfit, s="lambda.min") {
    ind <- which(coef(cvfit, s=s) != 0)
    df <- data.frame(
        feature=rownames(coef(cvfit, s=s))[ind],
        coeficient=coef(cvfit, s=s)[ind]
    )
    kable(df)
}

The function above uses the kable() function from knitr to produce a Markdown-ready table.

Answer 5

There is an approach with using coef() to glmnet() object (your model). In a case below index [[1]] indicate the number of outcome class in multinomial logistic regression, maybe for other models you shoould remove it.

coef_names_GLMnet <- coef(GLMnet, s = 0)[[1]]
row.names(coef_names_GLMnet)[coef_names_GLMnet@i+1]

row.names() indexes in such case needs incrementing (+1) because numeration of variables (data features) in coef() object begining from 0, but after transformation character vector numeration begining from 1.

Answer 6

Here, I wrote a reproducible example and fitted a binary (logistic) example using cv.glmnet. A glmnet model fit will also work. At the end of this example, I assembled non-zero coefficients, and associated features, into a data.frame called myResults:

library(glmnet)
X <- matrix(rnorm(100*10), 100, 10);
X[51:100, ] <- X[51:100, ] + 0.5; #artificially introduce difference in control cases
rownames(X) <- paste0("observation", 1:nrow(X));
colnames(X) <- paste0("feature",     1:ncol(X));

y <- factor( c(rep(1,50), rep(0,50)) ); #binary outcome class label
y
## [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
## [51] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
## Levels: 0 1

## Perform logistic model fit:
fit1 <- cv.glmnet(X, y, family="binomial", nfolds=5, type.measure="auc"); #with K-fold cross validation
# fit1 <- glmnet(X, y, family="binomial") #without cross validation also works

## Adapted from @Mehrad Mahmoudian:
myCoefs <- coef(fit1, s="lambda.min");
myCoefs[which(myCoefs != 0 ) ]               #coefficients: intercept included
## [1]  1.4945869 -0.6907010 -0.7578129 -1.1451275 -0.7494350 -0.3418030 -0.8012926 -0.6597648 -0.5555719
## [10] -1.1269725 -0.4375461
myCoefs@Dimnames[[1]][which(myCoefs != 0 ) ] #feature names: intercept included
## [1] "(Intercept)" "feature1"    "feature2"    "feature3"    "feature4"    "feature5"    "feature6"   
## [8] "feature7"    "feature8"    "feature9"    "feature10"  

## Asseble into a data.frame
myResults <- data.frame(
  features = myCoefs@Dimnames[[1]][ which(myCoefs != 0 ) ], #intercept included
  coefs    = myCoefs              [ which(myCoefs != 0 ) ]  #intercept included
)
myResults
##       features      coefs
## 1  (Intercept)  1.4945869
## 2     feature1 -0.6907010
## 3     feature2 -0.7578129
## 4     feature3 -1.1451275
## 5     feature4 -0.7494350
## 6     feature5 -0.3418030
## 7     feature6 -0.8012926
## 8     feature7 -0.6597648
## 9     feature8 -0.5555719
## 10    feature9 -1.1269725
## 11   feature10 -0.4375461

Answer 7

# requires tibble.
tidy_coef <- function(x){
    coef(x) %>%
    matrix %>%   # Coerce from sparse matrix to regular matrix.
    data.frame %>%  # Then dataframes.
    rownames_to_column %>%  # Add rownames as explicit variables.
    setNames(c("term","estimate"))
}

Without tibble:

tidy_coef2 <- function(x){
    x <- coef(x)
    data.frame(term=rownames(x),
               estimate=matrix(x)[,1],
               stringsAsFactors = FALSE)
}

Answer 8

Assuming you know how to obtain your lambda, I found two different ways to show the predictors needed in the selected model for that particular lambda. One of them includes the intercept. The lambda can be obtained using cross-validation by the mean of cv.glmnet from "glmnet" library. You might want to only look at the last lines for each method:

 myFittedLasso = glmnet(x=myXmatrix, y=myYresponse, family="binomial")
 myCrossValidated = cv.glmnet(x=myXmatrix, y=myYresponse, family="binomial")
 myLambda = myCrossValidated$lambda.1se  # can be simply lambda

 # Method 1 without the intercept
 myBetas = myFittedLasso$beta[, which(myFittedLasso$lambda == myLambda)]
 myBetas[myBetas != 0]
 ## myPredictor1    myPredictor2    myPredictor3
 ##   0.24289802      0.07561533      0.18299284


 # Method 2 with the intercept
 myCoefficients = coef(myFittedLasso, s=myLambda)
 dimnames(myCoefficients)[[1]][which(myCoefficients != 0)]
 ## [1] "(Intercept)"    "myPredictor1"    "M_myPredictor2"    "myPredictor3"

 myCoefficients[which(myCoefficients != 0)]
 ## [1] -4.07805560  0.24289802  0.07561533  0.18299284

Note that the example above implies a binomial distribution but the steps can be applied to any other kind.

Answer 9

I faced a similar issue when using glmnet from the tidymodels framework, where the model was trained within a workflow and neither coef() nor the above solutions worked.

What worked for me though, was part of the glmnet:::coef.glmnet code:

# taken from glmnet:::coef.glmnet
coefs <- predict(x, "lambda.min", type = "coefficients", exact = FALSE)

dd <- cbind(
  data.frame(var = rownames(coefs)),
  as.data.table(as.matrix(coefs))
)