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L1 Regularization

Last updatedUpdated: by Jakub Žovák · 1 min read

Properties
created 15.03.2026, 09:28
modified 26.07.2026, 13:37
published Empty
topics Regularization, Sparsity
authors Jakub
ai-assisted Yes

L1 regularization (also called Lasso in the statistics literature) adds the sum of absolute values of the weights to the training loss (absolute sum of weights + loss). It encourages the model to drive unimportant weights exactly to zero, producing sparse solutions ( Deep Learning, Ch. 7).

# Penalized Loss

$$ \mathcal{L}_{\text{L1}} = \mathcal{L} + \lambda \sum_{i} |w_i| $$

where:

  • \(\mathcal{L}\) is the original loss (e.g. cross-entropy)
  • \(\lambda > 0\) is the regularization strength
  • \(w_i\) are the model weights

# Gradient Update

$$ \frac{\partial \mathcal{L}_{\text{L1}}}{\partial w_i} = \frac{\partial \mathcal{L}}{\partial w_i} + \lambda \cdot \text{sign}(w_i) $$

The subgradient \(\text{sign}(w_i)\) applies a constant push toward zero, regardless of magnitude — so small weights are driven all the way to 0.

# Properties

  • Produces sparse weight vectors (many exactly-zero weights) — useful for feature selection
  • Non-differentiable at \(w_i = 0\); handled via subgradients or proximal operators
  • Less common in deep learning than L2; more common in linear models and sparse coding
  • Equivalent to placing a Laplace prior on weights from a Bayesian perspective

# Comparison with L2 Regularization

L1L2
Penalty(\sumw_i
EffectSparse (zeros out weights)Dense (shrinks all weights)
GeometryDiamond constraintSphere constraint
DifferentiabilityNo (at 0)Yes