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I am interested in solving the following problem

$$ \min_{\boldsymbol{\beta}} \left( \mathbf{y}-\mathbf{X}\boldsymbol{\beta} \right)^T W \left( \mathbf{y}-\mathbf{X}\boldsymbol{\beta} \right) + \lambda \left|\boldsymbol{\beta}\right|_1 $$

$\mathbf{y}$ is the vector of observations for each dataset.

$\mathbf{X}$ is the matrix of predictors.

$\boldsymbol{\beta}$ is the set of regression coefficients.

$\mathbf{W}$ is a diagonal matrix filled with positive real numbers.

using the LARS-Lasso approach. Is there any existing package that does this, meaning that it accepts weights for each as an input?

With all the normalizing of the datasets (centering and scaling) that need to be done, I am hesitant to pre-process my data and multiply both the observation and the predictors with $\sqrt{W}$ and feed it in to the algorithm.

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  • $\begingroup$ Are you asking for an R package? Are you wondering about anything other than the name of a package that would do this for you? If not, this question is probably best asked on the r-help-listserve; it would be off-topic here (please see our FAQ). $\endgroup$
    – gung - Reinstate Monica
    Nov 8, 2012 at 20:40
  • $\begingroup$ Nope. This question is not specific to R. I am interested in finding if there exists any implementation of lars-lasso that takes into accounts weights on datasets. $\endgroup$
    – Meenakshi
    Nov 8, 2012 at 20:56
  • $\begingroup$ You want to know if there's any implementation of these algorithms in any software anywhere? $\endgroup$
    – gung - Reinstate Monica
    Nov 8, 2012 at 21:18
  • $\begingroup$ I have seen several implementation on the internet, the lars package in R, the package in mlpack etc.. but I am not aware of any that takes weights for the datasets into account. $\endgroup$
    – Meenakshi
    Nov 9, 2012 at 2:04

1 Answer 1

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The glmnet Package solves the lasso problem using coordinate descent. It also provides features for adding in weights

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  • $\begingroup$ You can mark this reply as accepted (and change at a later time if another answer is proposed). $\endgroup$
    – chl
    May 5, 2013 at 14:19

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