Can L2 regularization be used for feature selection?
So while L2 regularization does not perform feature selection the same way as L1 does, it is more useful for feature *interpretation*: a predictive feature will get a non-zero coefficient, which is often not the case with L1.
Why is L2 better than L1 loss?
With outliers in the dataset, a L2(Loss function) tries to adjust the model according to these outliers on the expense of other good-samples, since the squared-error is going to be huge for these outliers(for error > 1). On the other hand L1(Least absolute deviation) is quite resistant to outliers.
What’s the difference between the L1 and L2?
Mathematically speaking, it adds a regularization term in order to prevent the coefficients to fit so perfectly to overfit. The difference between the L1 and L2 is just that L2 is the sum of the square of the weights, while L1 is just the sum of the weights. As follows: L1 regularization on least squares:
Which is more robust, L1 or L2 regularization?
L1 regularization is more robust than L2 regularization for a fairly obvious reason. L2 regularization takes the square of the weights, so the cost of outliers present in the data increases exponentially. L1 regularization takes the absolute values of the weights, so the cost only increases linearly. What solution has more possibilities?
How is L1 regularization used in feature selection?
L1 regularization does feature selection. It does this by assigning insignificant input features with zero weight and useful features with a non zero weight. In L1 regularization we penalize the absolute value of the weights. L1 regularization term is highlighted in the red box.
When to use sparse L1 or L2 norm?
When input features have weights closer to zero that leads to sparse L1 norm. In Sparse solution majority of the input features have zero weights and very few features have non zero weights. To predict ACT score not all input features have the same influence on the prediction.