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# -*- coding: utf-8 -*-
"""
LP solvers for optimal transport using cvxopt
"""
# Author: Remi Flamary <remi.flamary@unice.fr>
#
# License: MIT License
import numpy as np
import scipy as sp
import scipy.sparse as sps
import ot
try:
import cvxopt
from cvxopt import solvers, matrix, spmatrix
except ImportError:
cvxopt = False
def scipy_sparse_to_spmatrix(A):
"""Efficient conversion from scipy sparse matrix to cvxopt sparse matrix"""
coo = A.tocoo()
SP = spmatrix(coo.data.tolist(), coo.row.tolist(), coo.col.tolist(), size=A.shape)
return SP
def barycenter(A, M, weights=None, verbose=False, log=False, solver='interior-point'):
"""Compute the entropic regularized wasserstein barycenter of distributions A
The function solves the following optimization problem [16]:
.. math::
\mathbf{a} = arg\min_\mathbf{a} \sum_i W_{1}(\mathbf{a},\mathbf{a}_i)
where :
- :math:`W_1(\cdot,\cdot)` is the Wasserstein distance (see ot.emd.sinkhorn)
- :math:`\mathbf{a}_i` are training distributions in the columns of matrix :math:`\mathbf{A}`
The linear program is solved using the interior point solver from scipy.optimize.
If cvxopt solver if installed it can use cvxopt
Note that this problem do not scale well (both in memory and computational time).
Parameters
----------
A : np.ndarray (d,n)
n training distributions a_i of size d
M : np.ndarray (d,d)
loss matrix for OT
reg : float
Regularization term >0
weights : np.ndarray (n,)
Weights of each histogram a_i on the simplex (barycentric coodinates)
verbose : bool, optional
Print information along iterations
log : bool, optional
record log if True
solver : string, optional
the solver used, default 'interior-point' use the lp solver from
scipy.optimize. None, or 'glpk' or 'mosek' use the solver from cvxopt.
Returns
-------
a : (d,) ndarray
Wasserstein barycenter
log : dict
log dictionary return only if log==True in parameters
References
----------
.. [16] Agueh, M., & Carlier, G. (2011). Barycenters in the Wasserstein space. SIAM Journal on Mathematical Analysis, 43(2), 904-924.
"""
if weights is None:
weights = np.ones(A.shape[1]) / A.shape[1]
else:
assert(len(weights) == A.shape[1])
n_distributions = A.shape[1]
n = A.shape[0]
n2 = n * n
c = np.zeros((0))
b_eq1 = np.zeros((0))
for i in range(n_distributions):
c = np.concatenate((c, M.ravel() * weights[i]))
b_eq1 = np.concatenate((b_eq1, A[:, i]))
c = np.concatenate((c, np.zeros(n)))
lst_idiag1 = [sps.kron(sps.eye(n), np.ones((1, n))) for i in range(n_distributions)]
# row constraints
A_eq1 = sps.hstack((sps.block_diag(lst_idiag1), sps.coo_matrix((n_distributions * n, n))))
# columns constraints
lst_idiag2 = []
lst_eye = []
for i in range(n_distributions):
if i == 0:
lst_idiag2.append(sps.kron(np.ones((1, n)), sps.eye(n)))
lst_eye.append(-sps.eye(n))
else:
lst_idiag2.append(sps.kron(np.ones((1, n)), sps.eye(n - 1, n)))
lst_eye.append(-sps.eye(n - 1, n))
A_eq2 = sps.hstack((sps.block_diag(lst_idiag2), sps.vstack(lst_eye)))
b_eq2 = np.zeros((A_eq2.shape[0]))
# full problem
A_eq = sps.vstack((A_eq1, A_eq2))
b_eq = np.concatenate((b_eq1, b_eq2))
if not cvxopt or solver in ['interior-point']:
# cvxopt not installed or interior point
if solver is None:
solver = 'interior-point'
options = {'sparse': True, 'disp': verbose}
sol = sp.optimize.linprog(c, A_eq=A_eq, b_eq=b_eq, method=solver,
options=options)
x = sol.x
b = x[-n:]
else:
h = np.zeros((n_distributions * n2 + n))
G = -sps.eye(n_distributions * n2 + n)
sol = solvers.lp(matrix(c), scipy_sparse_to_spmatrix(G), matrix(h),
A=scipy_sparse_to_spmatrix(A_eq), b=matrix(b_eq),
solver=solver)
x = np.array(sol['x'])
b = x[-n:].ravel()
if log:
return b, sol
else:
return b
def free_support_barycenter(data_positions, data_weights, X_init, b_init, lamda, numItermax=100, stopThr=1e-5, verbose=False, log=False, **kwargs):
"""
Solves the free support (locations of the barycenters are optimized, not the weights) Wasserstein barycenter problem (i.e. the weighted Frechet mean for the 2-Wasserstein distance)
The function solves the Wasserstein barycenter problem when the barycenter measure is constrained to be supported on k atoms.
This problem is considered in [1] (Algorithm 2). There are two differences with the following codes:
- we do not optimize over the weights
- we do not do line search for the locations updates, we use i.e. theta = 1 in [1] (Algorithm 2). This can be seen as a discrete implementation of the fixed-point algorithm of [2] proposed in the continuous setting.
Parameters
----------
data_positions : list of (k_i,d) np.ndarray
The discrete support of a measure supported on k_i locations of a d-dimensional space (k_i can be different for each element of the list)
data_weights : list of (k_i,) np.ndarray
Numpy arrays where each numpy array has k_i non-negatives values summing to one representing the weights of each discrete input measure
X_init : (k,d) np.ndarray
Initialization of the support locations (on k atoms) of the barycenter
b_init : (k,) np.ndarray
Initialization of the weights of the barycenter (non-negatives, sum to 1)
lambda : (k,) np.ndarray
Initialization of the coefficients of the barycenter (non-negatives, sum to 1)
numItermax : int, optional
Max number of iterations
stopThr : float, optional
Stop threshol on error (>0)
verbose : bool, optional
Print information along iterations
log : bool, optional
record log if True
Returns
-------
X : (k,d) np.ndarray
Support locations (on k atoms) of the barycenter
References
----------
.. [1] Cuturi, Marco, and Arnaud Doucet. "Fast computation of Wasserstein barycenters." International Conference on Machine Learning. 2014.
.. [2] Álvarez-Esteban, Pedro C., et al. "A fixed-point approach to barycenters in Wasserstein space." Journal of Mathematical Analysis and Applications 441.2 (2016): 744-762.
"""
iter_count = 0
d = X_init.shape[1]
k = b_init.size
N = len(data_positions)
X = X_init
displacement_square_norm = 1e3
while ( displacement_square_norm > stopThr and iter_count < numItermax ):
T_sum = np.zeros((k, d))
for (data_positions_i, data_weights_i) in zip(data_positions, data_weights):
M_i = ot.dist(X, data_positions_i)
T_i = ot.emd(b_init, data_weights_i, M_i)
T_sum += np.reshape(1. / b_init, (-1, 1)) * np.matmul(T_i, data_positions_i)
X_previous = X
X = T_sum / N
displacement_square_norm = np.sum(np.square(X-X_previous))
iter_count += 1
return X
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