Sorry if this question comes across a little basic.

I am looking to use LASSO variable selection for a multiple linear regression model in R. I have 15 predictors, one of which is categorical(will that cause a problem?). After setting my $x$ and $y$ I use the following commands:

model = lars(x, y)

My problem is when I use coef(model). This returns a matrix with 15 rows, with one extra predictor added each time. However there is no suggestion as to which model to choose. Have I missed something? Is there a way I can get the lars package to return just one "best" model?

There are other posts suggesting using glmnet instead but this seems more complicated. An attempt is as follows, using the same $x$ and $y$. Have I missed something here?:

cv = cv.glmnet(x, y)
model = glmnet(x, y, type.gaussian="covariance", lambda=cv$lambda.min)
predict(model, type="coefficients")

The final command returns a list of my variables, the majority with a coefficient although some are =0. Is this the correct choice of the "best" model selected by LASSO? If I then fit a linear model with all my variables which had coefficients not=0 I get very similar, but slightly different, coefficient estimates. Is there a reason for this difference? Would it be acceptable to refit the linear model with these variables chosen by LASSO and take that as my final model? Otherwise I cannot see any p-values for significance. Have I missed anything?



ensure that that glmnet uses multiple linear regression?

Does the automatic normalisation of the variables affect the coefficients at all? Is there any way to include interaction terms in a LASSO procedure?

I am looking to use this procedure more as a demonstration of how LASSO can be used than for any model that will actually be used for any important inference/prediction if that changes anything.

Thank you for taking the time to read this. Any general comments on LASSO/lars/glmnet would also be greatly appreciated.

  • 4
    $\begingroup$ As a side comment, if you want to interpret the result be sure to demonstrate the that set of variables selected by lasso is stable. This can be done using Monte Carlo simulation or by bootstrapping your own dataset. $\endgroup$ – Frank Harrell Sep 15 '13 at 8:43

Using glmnet is really easy once you get the grasp of it thanks to its excellent vignette in http://web.stanford.edu/~hastie/glmnet/glmnet_alpha.html (you can also check the CRAN package page). As for the best lambda for glmnet, the rule of thumb is to use

cvfit <- glmnet::cv.glmnet(x, y)
coef(cvfit, s = "lambda.1se")

instead of lambda.min.

To do the same for lars you have to do it by hand. Here is my solution

cv <- lars::cv.lars(x, y, plot.it = FALSE, mode = "step")
idx <- which.max(cv$cv - cv$cv.error <= min(cv$cv))
coef(lars::lars(x, y))[idx,]

Bear in mind that this is not exactly the same, because this is stopping at a lasso knot (when a variable enters) instead of at any point.

Please note that glmnet is the preferred package now, it is actively maintained, more so than lars, and that there have been questions about glmnet vs lars answered before (algorithms used differ).

As for your question of using lasso to choose variables and then fit OLS, it is an ongoing debate. Google for OLS post Lasso and there are some papers discussing the topic. Even the authors of Elements of Statistical Learning admit it is possible.

Edit: Here is the code to reproduce more accurately what glmnet does in lars

  cv <- lars::cv.lars(x, y, plot.it = FALSE)
  ideal_l1_ratio <- cv$index[which.max(cv$cv - cv$cv.error <= min(cv$cv))]
  obj <- lars::lars(x, y)
  scaled_coefs <- scale(obj$beta, FALSE, 1 / obj$normx)
  l1 <- apply(X = scaled_coefs, MARGIN = 1, FUN = function(x) sum(abs(x)))
  coef(obj)[which.max(l1 / tail(l1, 1) > ideal_l1_ratio),]
  • $\begingroup$ +1 Great answer! Could you or anyone possibly elaborate on why lambda.1se is the rule of thumb, instead of lambda.min? $\endgroup$ – Erosennin Jan 23 '19 at 12:00
  • $\begingroup$ After 4 years of writing this (and not having used lasso for a while) my memory just vanished. Sorry! $\endgroup$ – Juancentro Jan 23 '19 at 12:04
  • $\begingroup$ the 1se , gives the most regularized model ( how far can Lamba be pushed) and the CV mse(or whatever measurement you chose) is within 1 se of min(MSE) you got at. $\endgroup$ – Yash Jul 17 '20 at 14:19

I'm returning to this question from a while ago since I think I've solved the correct solution.

Here's a replica using the mtcars dataset:




variables<-variables[variables %ni% '(Intercept)']

'variables' gives you the list of the variables that solve the best solution.

  • 1
    $\begingroup$ I was looking to the code and I find that "testing" was not defined yet and therefore the code: "final.list<-testing[-removed] #removing variables" gives the error: object not found So looking to the code I suppose that instead of using "testing" it should be used "cp.list" so that the code will be: final.list<-cp.list[-removed] #removing variables final.list<-c(final.list,duplicates) #adding in those vars which were both removed then added later Let me know if this is correct Kind regards $\endgroup$ – user55101 Sep 2 '14 at 14:27
  • 4
    $\begingroup$ `%ni%`<-Negate(`%ni%`); ##looks wrong. While `%ni%`<-Negate(`%in%`); ##looks right. I think the stackexchange formatter messed it up... $\endgroup$ – Chris Jun 30 '15 at 4:17
  • $\begingroup$ Can you elaborate how you chose the nfolds=5 and alpha=0.5 parameters? $\endgroup$ – colin Dec 19 '18 at 22:08

Perhaps the comparison with forward selection stepwise regression will help (see the following link to a site by one of the authors http://www-stat.stanford.edu/~tibs/lasso/simple.html). This is the approach used in Chapter 3.4.4 of The Elements of Statistical Learning (available online for free). I thought that Chapter 3.6 in that book helped to understand the relationship between least squares, best subset, and lasso (plus a couple of other procedures). I also find it helpful to take the transpose of the coefficient, t(coef(model)) and write.csv, so that I can open it in Excel along with a copy of the plot(model) on the side. You might want to sort by the last column, which contains the least squares estimate. Then you can see clearly how each variable gets added at each piecewise step and how the coefficients change as a result. Of course this is not the whole story, but hopefully it will be a start.


lars and glmnet operate on raw matrices. To includ interaction terms, you will have to construct the matrices yourself. That means one column per interaction (which is per level per factor if you have factors). Look into lm() to see how it does it (warning: there be dragons).

To do it right now, do something like: To make an interaction term manually, you could (but maybe shouldn't, because it's slow) do:

int = D["x1"]*D["x2"]
names(int) = c("x1*x2")
D = cbind(D, int)

Then to use this in lars (assuming you have a y kicking around):

lars(as.matrix(D), as.matrix(y))

I wish I could help you more with the other questions. I found this one because lars is giving me grief and the documentation in it and on the web is very thin.

  • 2
    $\begingroup$ "Warning: there be dragons" This is pretty easy with model.matrix(). $\endgroup$ – Gregor Thomas Nov 25 '14 at 23:52

LARS solves the ENTIRE solution path. The solution path is piecewise linear -- there are a finite number of "notch" points (i.e., values of the regularization parameter) at which the solution changes.

So the matrix of solutions you're getting is all the possible solutions. In the list that it returns, it should also give you the values of the regularization parameter.

  • $\begingroup$ Thank you for your answer. Is there a way to display the values of the regularisation parameter? Additionally is there a way to then choose between the solutions based on this parameter? (Also is the parameter lambda?) $\endgroup$ – James May 9 '13 at 10:31
  • $\begingroup$ Note that piecewise-linearity does not mean that the lines are horizontal, and thus the solution is changing all the time with lambda. For example, for predictive purposes one would have a grid of lambda values not only at but also between the knots. It is quite possible that some point between the knots yields the best better predictive performance. $\endgroup$ – Richard Hardy Aug 9 '16 at 16:09

Not the answer you're looking for? Browse other questions tagged or ask your own question.