Compute Euclidean projections on the simplex or L1-ball

simplex_projection.py

""" Module to compute projections on the positive simplex or the L1-ball

A positive simplex is a set X = { \mathbf{x} | \sum_i x_i = s, x_i \geq 0 }

The (unit) L1-ball is the set X = { \mathbf{x} | || x ||_1 \leq 1 }

Adrien Gaidon - INRIA - 2011
"""


import numpy as np


def euclidean_proj_simplex(v, s=1):
    """ Compute the Euclidean projection on a positive simplex

    Solves the optimisation problem (using the algorithm from [1]):

        min_w 0.5 * || w - v ||_2^2 , s.t. \sum_i w_i = s, w_i >= 0 

    Parameters
    ----------
    v: (n,) numpy array,
       n-dimensional vector to project

    s: int, optional, default: 1,
       radius of the simplex

    Returns
    -------
    w: (n,) numpy array,
       Euclidean projection of v on the simplex

    Notes
    -----
    The complexity of this algorithm is in O(n log(n)) as it involves sorting v.
    Better alternatives exist for high-dimensional sparse vectors (cf. [1])
    However, this implementation still easily scales to millions of dimensions.

    References
    ----------
    [1] Efficient Projections onto the .1-Ball for Learning in High Dimensions
        John Duchi, Shai Shalev-Shwartz, Yoram Singer, and Tushar Chandra.
        International Conference on Machine Learning (ICML 2008)
        http://www.cs.berkeley.edu/~jduchi/projects/DuchiSiShCh08.pdf
    """
    assert s > 0, "Radius s must be strictly positive (%d <= 0)" % s
    n, = v.shape  # will raise ValueError if v is not 1-D
    # check if we are already on the simplex
    if v.sum() == s and np.alltrue(v >= 0):
        # best projection: itself!
        return v
    # get the array of cumulative sums of a sorted (decreasing) copy of v
    u = np.sort(v)[::-1]
    cssv = np.cumsum(u)
    # get the number of > 0 components of the optimal solution
    rho = np.nonzero(u * np.arange(1, n+1) > (cssv - s))[0][-1]
    # compute the Lagrange multiplier associated to the simplex constraint
    theta = (cssv[rho] - s) / (rho + 1.0)
    # compute the projection by thresholding v using theta
    w = (v - theta).clip(min=0)
    return w


def euclidean_proj_l1ball(v, s=1):
    """ Compute the Euclidean projection on a L1-ball

    Solves the optimisation problem (using the algorithm from [1]):

        min_w 0.5 * || w - v ||_2^2 , s.t. || w ||_1 <= s

    Parameters
    ----------
    v: (n,) numpy array,
       n-dimensional vector to project

    s: int, optional, default: 1,
       radius of the L1-ball

    Returns
    -------
    w: (n,) numpy array,
       Euclidean projection of v on the L1-ball of radius s

    Notes
    -----
    Solves the problem by a reduction to the positive simplex case

    See also
    --------
    euclidean_proj_simplex
    """
    assert s > 0, "Radius s must be strictly positive (%d <= 0)" % s
    n, = v.shape  # will raise ValueError if v is not 1-D
    # compute the vector of absolute values
    u = np.abs(v)
    # check if v is already a solution
    if u.sum() <= s:
        # L1-norm is <= s
        return v
    # v is not already a solution: optimum lies on the boundary (norm == s)
    # project *u* on the simplex
    w = euclidean_proj_simplex(u, s=s)
    # compute the solution to the original problem on v
    w *= np.sign(v)
    return w

I think there is an error on line 59, it should be
theta = float(cssv[rho] - s) / (rho+1)

I think there is an error on line 59, it should be
theta = float(cssv[rho] - s) / (rho+1)

Thanks @omarfoq : indeed, testing on np.array([2, 2]) with the original gist returns array([0., 0.]), but with your modification, at least in this case, it returns the right result (array([0.5, 0.5]))