I am working with the famous white wine dataset, and I am trying to fit a logistic model on it, where I also perform ridge regression on this logistic model. Finally, I want to calculate the test error of this model using 10 fold CV. Does this make sense? I ask because trying to do this with R seems to not be easily possible. Here is my code -

# Data pre-processing
winedata <- read.delim("winequality-white.csv", sep = ';')
winedata$quality[winedata$quality< 7] <- "0" #recode
winedata$quality[winedata$quality>=7] <- "1" #recode
winedata$quality <- factor(winedata$quality)# Convert the column to a factor
names(winedata)[names(winedata) == "quality"] <- "good"      #rename 'quality' to 'good'

# I then used 10 fold C to find the best value for lambda for Ridge regression, not adding the code for that here
ridge.model <- glmnet(x, y, alpha = 0, family = "binomial", lambda = bestlam) # fit logistic model with ridge regression. This is the model I want finally
ridge.model.cv.err<- cv.glmnet(x,y, lambda = bestlam, cost1, K=10) # trying to calculate the 10 fold CV of the final model. THIS GIVES ERROR
ridge.model.cv.err$delta # this is what I hoped would give me the test error

This gives the error - Error in cv.glmnet(x, y, lambda = bestlam, cost1, K = 10) : Need more than one value of lambda for cv.glmnet

Am I doing something very wrong here, something that doesn't make sense?

  • $\begingroup$ what does it mean to perform ridge regression on the logistic model? Is it L2 regularization over the logistic model that you've meant? $\endgroup$
    – gunes
    Apr 4, 2020 at 20:24
  • $\begingroup$ @gunes, I think it would be thought as penalized logistic regression, where I specify the lambda value that specifies the coefficient shrinkage. I calculated this optimum lambda value by doing 10 fold CV for ridge regression on the data $\endgroup$
    – user1995
    Apr 4, 2020 at 20:34
  • $\begingroup$ I read about Penalized Logistic Regression here - sthda.com/english/articles/36-classification-methods-essentials/… $\endgroup$
    – user1995
    Apr 4, 2020 at 20:40
  • $\begingroup$ I think it’s telling you to give all ten $\lambda$ values. What happens if you give it rep(lambda, 10)? $\endgroup$
    – Dave
    Apr 4, 2020 at 21:07
  • $\begingroup$ @Dave number of folds is irrelevant to the number of different lambda values $\endgroup$
    – gunes
    Apr 4, 2020 at 21:24

1 Answer 1


Finding best $\lambda$ from Ridge Regression and using it in logistic regression is not a good idea. Optimisation objectives significantly differ between the two models. You need to find the best $\lambda$ for the penalised logreg.

The error in your code is actually a good warning. You want to do a cross-validation but there is nothing to cross-validate it says.

  • $\begingroup$ Thanks for the answer! I understand that finding the best lambda should be done after applying ridge regression to the logistic model, and I will surely do it. But I mainly wanted to ask if cross validation can then be applied to the penalized logistic regression model to get the test error. I am more interested in doing CV for the test error than for finding the best lambda. I have framed the problem from a code point of view here as well - stackoverflow.com/questions/61034771/… $\endgroup$
    – user1995
    Apr 4, 2020 at 21:16
  • 1
    $\begingroup$ Maybe you can set lambda = c(1,1+1e-12) and get your error. Such a small deviance in lambda wouldn't produce a different error. This is hacky, and there are probably better ways to do it (other than coding the cv yourself), but I'm not extremely familiar with glmnet. $\endgroup$
    – gunes
    Apr 4, 2020 at 21:19
  • $\begingroup$ Thanks @gunes. I tried that but the error it gave me using the cvm output of cv.glmnet was much higher than what I have got for this dataset using best subset selection and other methods. I am wondering if cv.glmnet can even be used to get CV test error! Maybe it is only to be used for finding the best lambda parameter $\endgroup$
    – user1995
    Apr 4, 2020 at 21:39
  • $\begingroup$ maybe your best lambda isn‘t the best as I warned you. I don’t think there is any reason for it to give wrong errors. $\endgroup$
    – gunes
    Apr 4, 2020 at 21:42

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