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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
# import compiled emd
from .emd_wrap import emd_c, checkResult
from ..utils import parmap
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
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, resultCode = emd_c(a, b, M, numItermax)
resultCodeString = checkResult(resultCode)
if log:
log = {}
log['cost'] = cost
log['u'] = u
log['v'] = v
log['warning'] = resultCodeString
log['resultCode'] = resultCode
return G, log
return G
def emd2(a, b, M, processes=multiprocessing.cpu_count(), numItermax=100000, log=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.
Returns
-------
gamma: (ns x nt) ndarray
Optimal transportation matrix for the given parameters
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:
def f(b):
G, cost, u, v, resultCode = emd_c(a, b, M, numItermax)
resultCodeString = checkResult(resultCode)
log = {}
log['G'] = G
log['u'] = u
log['v'] = v
log['warning'] = resultCodeString
log['resultCode'] = resultCode
return [cost, log]
else:
def f(b):
G, cost, u, v, resultCode = emd_c(a, b, M, numItermax)
checkResult(resultCode)
return cost
if len(b.shape) == 1:
return f(b)
nb = b.shape[1]
# res = [emd2_c(a, b[:, i].copy(), M, numItermax) for i in range(nb)]
res = parmap(f, [b[:, i] for i in range(nb)], processes)
return res
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