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# -*- coding: utf-8 -*-
"""
Solvers for the original linear program OT problem
"""
# Author: Remi Flamary <remi.flamary@unice.fr>
#
# License: MIT License
import multiprocessing
import numpy as np
from .import cvx
# import compiled emd
from .emd_wrap import emd_c, check_result
from ..utils import parmap
from .cvx import barycenter
from ..utils import dist
__all__=['emd', 'emd2', 'barycenter', 'free_support_barycenter', 'cvx']
def emd(a, b, M, numItermax=100000, log=False):
"""Solves the Earth Movers distance problem and returns the OT matrix
.. math::
\gamma = arg\min_\gamma <\gamma,M>_F
s.t. \gamma 1 = a
\gamma^T 1= b
\gamma\geq 0
where :
- M is the metric cost matrix
- a and b are the sample weights
Uses the algorithm proposed in [1]_
Parameters
----------
a : (ns,) ndarray, float64
Source histogram (uniform weigth if empty list)
b : (nt,) ndarray, float64
Target histogram (uniform weigth if empty list)
M : (ns,nt) ndarray, float64
loss matrix
numItermax : int, optional (default=100000)
The maximum number of iterations before stopping the optimization
algorithm if it has not converged.
log: boolean, optional (default=False)
If True, returns a dictionary containing the cost and dual
variables. Otherwise returns only the optimal transportation matrix.
Returns
-------
gamma: (ns x nt) ndarray
Optimal transportation matrix for the given parameters
log: dict
If input log is true, a dictionary containing the cost and dual
variables and exit status
Examples
--------
Simple example with obvious solution. The function emd accepts lists and
perform automatic conversion to numpy arrays
>>> import ot
>>> a=[.5,.5]
>>> b=[.5,.5]
>>> M=[[0.,1.],[1.,0.]]
>>> ot.emd(a,b,M)
array([[ 0.5, 0. ],
[ 0. , 0.5]])
References
----------
.. [1] Bonneel, N., Van De Panne, M., Paris, S., & Heidrich, W.
(2011, December). Displacement interpolation using Lagrangian mass
transport. In ACM Transactions on Graphics (TOG) (Vol. 30, No. 6, p.
158). ACM.
See Also
--------
ot.bregman.sinkhorn : Entropic regularized OT
ot.optim.cg : General regularized OT"""
a = np.asarray(a, dtype=np.float64)
b = np.asarray(b, dtype=np.float64)
M = np.asarray(M, dtype=np.float64)
# if empty array given then use unifor distributions
if len(a) == 0:
a = np.ones((M.shape[0],), dtype=np.float64) / M.shape[0]
if len(b) == 0:
b = np.ones((M.shape[1],), dtype=np.float64) / M.shape[1]
G, cost, u, v, result_code = emd_c(a, b, M, numItermax)
result_code_string = check_result(result_code)
if log:
log = {}
log['cost'] = cost
log['u'] = u
log['v'] = v
log['warning'] = result_code_string
log['result_code'] = result_code
return G, log
return G
def emd2(a, b, M, processes=multiprocessing.cpu_count(),
numItermax=100000, log=False, return_matrix=False):
"""Solves the Earth Movers distance problem and returns the loss
.. math::
\gamma = arg\min_\gamma <\gamma,M>_F
s.t. \gamma 1 = a
\gamma^T 1= b
\gamma\geq 0
where :
- M is the metric cost matrix
- a and b are the sample weights
Uses the algorithm proposed in [1]_
Parameters
----------
a : (ns,) ndarray, float64
Source histogram (uniform weigth if empty list)
b : (nt,) ndarray, float64
Target histogram (uniform weigth if empty list)
M : (ns,nt) ndarray, float64
loss matrix
numItermax : int, optional (default=100000)
The maximum number of iterations before stopping the optimization
algorithm if it has not converged.
log: boolean, optional (default=False)
If True, returns a dictionary containing the cost and dual
variables. Otherwise returns only the optimal transportation cost.
return_matrix: boolean, optional (default=False)
If True, returns the optimal transportation matrix in the log.
Returns
-------
gamma: (ns x nt) ndarray
Optimal transportation matrix for the given parameters
log: dict
If input log is true, a dictionary containing the cost and dual
variables and exit status
Examples
--------
Simple example with obvious solution. The function emd accepts lists and
perform automatic conversion to numpy arrays
>>> import ot
>>> a=[.5,.5]
>>> b=[.5,.5]
>>> M=[[0.,1.],[1.,0.]]
>>> ot.emd2(a,b,M)
0.0
References
----------
.. [1] Bonneel, N., Van De Panne, M., Paris, S., & Heidrich, W.
(2011, December). Displacement interpolation using Lagrangian mass
transport. In ACM Transactions on Graphics (TOG) (Vol. 30, No. 6, p.
158). ACM.
See Also
--------
ot.bregman.sinkhorn : Entropic regularized OT
ot.optim.cg : General regularized OT"""
a = np.asarray(a, dtype=np.float64)
b = np.asarray(b, dtype=np.float64)
M = np.asarray(M, dtype=np.float64)
# if empty array given then use unifor distributions
if len(a) == 0:
a = np.ones((M.shape[0],), dtype=np.float64) / M.shape[0]
if len(b) == 0:
b = np.ones((M.shape[1],), dtype=np.float64) / M.shape[1]
if log or return_matrix:
def f(b):
G, cost, u, v, resultCode = emd_c(a, b, M, numItermax)
result_code_string = check_result(resultCode)
log = {}
if return_matrix:
log['G'] = G
log['u'] = u
log['v'] = v
log['warning'] = result_code_string
log['result_code'] = resultCode
return [cost, log]
else:
def f(b):
G, cost, u, v, result_code = emd_c(a, b, M, numItermax)
check_result(result_code)
return cost
if len(b.shape) == 1:
return f(b)
nb = b.shape[1]
res = parmap(f, [b[:, i] for i in range(nb)], processes)
return res
def free_support_barycenter(measures_locations, measures_weights, X_init, b=None, weights=None, numItermax=100, stopThr=1e-7, verbose=False, log=None):
"""
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
----------
measures_locations : 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)
measures_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 : (k,) np.ndarray
Initialization of the weights of the barycenter (non-negatives, sum to 1)
weights : (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
N = len(measures_locations)
k = X_init.shape[0]
d = X_init.shape[1]
if b is None:
b = np.ones((k,))/k
if weights is None:
weights = np.ones((N,)) / N
X = X_init
log_dict = {}
displacement_square_norms = []
displacement_square_norm = stopThr + 1.
while ( displacement_square_norm > stopThr and iter_count < numItermax ):
T_sum = np.zeros((k, d))
for (measure_locations_i, measure_weights_i, weight_i) in zip(measures_locations, measures_weights, weights.tolist()):
M_i = dist(X, measure_locations_i)
T_i = emd(b, measure_weights_i, M_i)
T_sum = T_sum + weight_i * np.reshape(1. / b, (-1, 1)) * np.matmul(T_i, measure_locations_i)
displacement_square_norm = np.sum(np.square(T_sum-X))
if log:
displacement_square_norms.append(displacement_square_norm)
X = T_sum
if verbose:
print('iteration %d, displacement_square_norm=%f\n', iter_count, displacement_square_norm)
iter_count += 1
if log:
log_dict['displacement_square_norms'] = displacement_square_norms
return X, log_dict
else:
return X
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