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# -*- coding: utf-8 -*-
"""
Bregman projections for regularized OT
"""
import numpy as np
def sinkhorn(a,b, M, reg,numItermax = 1000,stopThr=1e-9,verbose=False,log=False):
"""
Solve the entropic regularization optimal transport problem and return the OT matrix
The function solves the following optimization problem:
.. math::
\gamma = arg\min_\gamma <\gamma,M>_F + reg\cdot\Omega(\gamma)
s.t. \gamma 1 = a
\gamma^T 1= b
\gamma\geq 0
where :
- M is the (ns,nt) metric cost matrix
- :math:`\Omega` is the entropic regularization term :math:`\Omega(\gamma)=\sum_{i,j} \gamma_{i,j}\log(\gamma_{i,j})`
- a and b are source and target weights (sum to 1)
The algorithm used for solving the problem is the Sinkhorn-Knopp matrix scaling algorithm as proposed in [2]_
Parameters
----------
a : np.ndarray (ns,)
samples weights in the source domain
b : np.ndarray (nt,)
samples in the target domain
M : np.ndarray (ns,nt)
loss matrix
reg: float
Regularization term >0
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
-------
gamma: (ns x nt) ndarray
Optimal transportation matrix for the given parameters
log: dict
log dictionary return only if log==True in parameters
Examples
--------
>>> a=[.5,.5]
>>> b=[.5,.5]
>>> M=[[0.,1.],[1.,0.]]
>>> ot.sinkhorn(a,b,M,1)
array([[ 0.36552929, 0.13447071],
[ 0.13447071, 0.36552929]])
References
----------
.. [2] M. Cuturi, Sinkhorn Distances : Lightspeed Computation of Optimal Transport, Advances in Neural Information Processing Systems (NIPS) 26, 2013
See Also
--------
ot.lp.emd : Unregularized 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 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]
# init data
Nini = len(a)
Nfin = len(b)
cpt = 0
if log:
log={'err':[]}
# we assume that no distances are null except those of the diagonal of distances
u = np.ones(Nini)/Nini
v = np.ones(Nfin)/Nfin
uprev=np.zeros(Nini)
vprev=np.zeros(Nini)
#print reg
K = np.exp(-M/reg)
#print np.min(K)
Kp = np.dot(np.diag(1/a),K)
transp = K
cpt = 0
err=1
while (err>stopThr and cpt<numItermax):
if np.any(np.dot(K.T,u)==0) or np.any(np.isnan(u)) or np.any(np.isnan(v)):
# we have reached the machine precision
# come back to previous solution and quit loop
print('Warning: numerical errrors')
if cpt!=0:
u = uprev
v = vprev
break
uprev = u
vprev = v
v = np.divide(b,np.dot(K.T,u))
u = 1./np.dot(Kp,v)
if cpt%10==0:
# we can speed up the process by checking for the error only all the 10th iterations
transp = np.dot(np.diag(u),np.dot(K,np.diag(v)))
err = np.linalg.norm((np.sum(transp,axis=0)-b))**2
if log:
log['err'].append(err)
if verbose:
if cpt%200 ==0:
print('{:5s}|{:12s}'.format('It.','Err')+'\n'+'-'*19)
print('{:5d}|{:8e}|'.format(cpt,err))
cpt = cpt +1
#print 'err=',err,' cpt=',cpt
if log:
return np.dot(np.diag(u),np.dot(K,np.diag(v))),log
else:
return np.dot(np.diag(u),np.dot(K,np.diag(v)))
def geometricBar(weights,alldistribT):
"""return the weighted geometric mean of distributions"""
assert(len(weights)==alldistribT.shape[1])
return np.exp(np.dot(np.log(alldistribT),weights.T))
def geometricMean(alldistribT):
"""return the geometric mean of distributions"""
return np.exp(np.mean(np.log(alldistribT),axis=1))
def projR(gamma,p):
#return np.dot(np.diag(p/np.maximum(np.sum(gamma,axis=1),1e-10)),gamma)
return np.multiply(gamma.T,p/np.maximum(np.sum(gamma,axis=1),1e-10)).T
def projC(gamma,q):
#return (np.dot(np.diag(q/np.maximum(np.sum(gamma,axis=0),1e-10)),gamma.T)).T
return np.multiply(gamma,q/np.maximum(np.sum(gamma,axis=0),1e-10))
def barycenter(A,M,reg, weights=None, numItermax = 1000, stopThr=1e-4,verbose=False,log=False):
"""Compute the entropic regularized wasserstein barycenter of distributions A
The function solves the following optimization problem:
.. math::
\mathbf{a} = arg\min_\mathbf{a} \sum_i W_{reg}(\mathbf{a},\mathbf{a}_i)
where :
- :math:`W_{reg}(\cdot,\cdot)` is the entropic regularized Wasserstein distance (see ot.bregman.sinkhorn)
- :math:`\mathbf{a}_i` are training distributions in the columns of matrix :math:`\mathbf{A}`
- reg and :math:`\mathbf{M}` are respectively the regularization term and the cost matrix for OT
The algorithm used for solving the problem is the Sinkhorn-Knopp matrix scaling algorithm as proposed in [3]_
Parameters
----------
A : np.ndarray (d,n)
n training distributions of size d
M : np.ndarray (d,d)
loss matrix for OT
reg: float
Regularization term >0
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
-------
a: (d,) ndarray
Wasserstein barycenter
log: dict
log dictionary return only if log==True in parameters
References
----------
.. [3] Benamou, J. D., Carlier, G., Cuturi, M., Nenna, L., & Peyré, G. (2015). Iterative Bregman projections for regularized transportation problems. SIAM Journal on Scientific Computing, 37(2), A1111-A1138.
"""
if weights is None:
weights=np.ones(A.shape[1])/A.shape[1]
else:
assert(len(weights)==A.shape[1])
if log:
log={'err':[]}
#M = M/np.median(M) # suggested by G. Peyre
K = np.exp(-M/reg)
cpt = 0
err=1
UKv=np.dot(K,np.divide(A.T,np.sum(K,axis=0)).T)
u = (geometricMean(UKv)/UKv.T).T
while (err>stopThr and cpt<numItermax):
cpt = cpt +1
UKv=u*np.dot(K,np.divide(A,np.dot(K,u)))
u = (u.T*geometricBar(weights,UKv)).T/UKv
if cpt%10==1:
err=np.sum(np.std(UKv,axis=1))
# log and verbose print
if log:
log['err'].append(err)
if verbose:
if cpt%200 ==0:
print('{:5s}|{:12s}'.format('It.','Err')+'\n'+'-'*19)
print('{:5d}|{:8e}|'.format(cpt,err))
if log:
log['niter']=cpt
return geometricBar(weights,UKv),log
else:
return geometricBar(weights,UKv)
def unmix(a,D,M,M0,h0,reg,reg0,alpha,numItermax = 1000, stopThr=1e-3,verbose=False,log=False):
"""
Compute the unmixing of an observation with a given dictionary using Wasserstein distance
The function solve the following optimization problem:
.. math::
\mathbf{h} = arg\min_\mathbf{h} (1- \\alpha) W_{M,reg}(\mathbf{a},\mathbf{Dh})+\\alpha W_{M0,reg0}(\mathbf{h}_0,\mathbf{h})
where :
- :math:`W_{M,reg}(\cdot,\cdot)` is the entropic regularized Wasserstein distance with M loss matrix (see ot.bregman.sinkhorn)
- :math:`\mathbf{a}` is an observed distribution, :math:`\mathbf{h}_0` is aprior on unmixing
- reg and :math:`\mathbf{M}` are respectively the regularization term and the cost matrix for OT data fitting
- reg0 and :math:`\mathbf{M0}` are respectively the regularization term and the cost matrix for regularization
- :math:`\\alpha`weight data fitting and regularization
The optimization problem is solved suing the algorithm described in [4]
distrib : distribution to unmix
D : Dictionnary
M : Metric matrix in the space of the distributions to unmix
M0 : Metric matrix in the space of the 'abundance values' to solve for
h0 : prior on solution (generally uniform distribution)
reg,reg0 : transport regularizations
alpha : how much should we trust the prior ? ([0,1])
Parameters
----------
a : np.ndarray (d)
observed distribution
D : np.ndarray (d,n)
dictionary matrix
M : np.ndarray (d,d)
loss matrix
M0 : np.ndarray (n,n)
loss matrix
h0 : np.ndarray (n,)
prior on h
reg: float
Regularization term >0 (Wasserstein data fitting)
reg0: float
Regularization term >0 (Wasserstein reg with h0)
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
-------
a: (d,) ndarray
Wasserstein barycenter
log: dict
log dictionary return only if log==True in parameters
References
----------
.. [4] S. Nakhostin, N. Courty, R. Flamary, D. Tuia, T. Corpetti, Supervised planetary unmixing with optimal transport, Whorkshop on Hyperspectral Image and Signal Processing : Evolution in Remote Sensing (WHISPERS), 2016.
"""
#M = M/np.median(M)
K = np.exp(-M/reg)
#M0 = M0/np.median(M0)
K0 = np.exp(-M0/reg0)
old = h0
err=1
cpt=0
#log = {'niter':0, 'all_err':[]}
if log:
log={'err':[]}
while (err>stopThr and cpt<numItermax):
K = projC(K,a)
K0 = projC(K0,h0)
new = np.sum(K0,axis=1)
inv_new = np.dot(D,new) # we recombine the current selection from dictionnary
other = np.sum(K,axis=1)
delta = np.exp(alpha*np.log(other)+(1-alpha)*np.log(inv_new)) # geometric interpolation
K = projR(K,delta)
K0 = np.dot(np.diag(np.dot(D.T,delta/inv_new)),K0)
err=np.linalg.norm(np.sum(K0,axis=1)-old)
old = new
if log:
log['err'].append(err)
if verbose:
if cpt%200 ==0:
print('{:5s}|{:12s}'.format('It.','Err')+'\n'+'-'*19)
print('{:5d}|{:8e}|'.format(cpt,err))
cpt = cpt+1
if log:
log['niter']=cpt
return np.sum(K0,axis=1),log
else:
return np.sum(K0,axis=1)
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