Machine Learning · medium

Explain L1 vs L2 regularization.

Asked in ML Engineer interviews, in the Machine Learning round.

Short answer

L1 (Lasso) drives weights to zero (sparsity/selection); L2 (Ridge) shrinks weights smoothly.

How to answer it

Both add a penalty on the weights to the loss, so the model prefers small coefficients unless the data insists. They differ in the shape of the penalty, and that shape decides what happens to weights near zero.

• L2 (ridge) penalises the sum of squared weights. The gradient shrinks each weight in proportion to its size, so weights get small but almost never exactly zero. All features stay in.
• L1 (lasso) penalises the sum of absolute weights. The gradient is constant in magnitude, so small weights are pushed all the way to zero. It selects features.

The geometric picture, if they want it: the L1 constraint region is a diamond with corners on the axes, and the loss contours tend to touch it at a corner, where some coordinates are zero; the L2 region is a circle with no corners.

from sklearn.linear_model import Lasso, Ridge
Lasso(alpha=0.01).fit(X, y).coef_    # many exact zeros
Ridge(alpha=1.0).fit(X, y).coef_     # all small, none zero

When to use which: L1 when you have many features and expect few to matter, or need an interpretable model; L2 when features are correlated (L1 picks one of a correlated pair arbitrarily, L2 shares the weight) or when you just want stability. Elastic net mixes both. Scale the features first; a penalty on raw weights punishes whichever feature happens to be measured in small units.

• What they are checking: "L1 gives zeros, L2 shrinks", and why.
• Common mistake: skipping feature scaling, so the regularisation strength means something different for every column.

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