1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
|
# -*- coding: utf-8 -*-
"""
Bregman projection for regularized Otimal transport
"""
import numpy as np
def sinkhorn(a,b, M, reg,numItermax = 1000,stopThr=1e-9):
"""
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 [1]_
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
Returns
-------
gamma: (ns x nt) ndarray
Optimal transportation matrix for the given parameters
References
----------
.. [1] M. Cuturi, Sinkhorn Distances : Lightspeed Computation of Optimal Transport, Advances in Neural Information Processing Systems (NIPS) 26, 2013
See Also
--------
ot.emd.emd : Unregularized optimal ransport
"""
# init data
Nini = len(a)
Nfin = len(b)
cpt = 0
# 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
cpt = cpt +1
#print 'err=',err,' cpt=',cpt
return np.dot(np.diag(u),np.dot(K,np.diag(v)))
def geometricBar(weights,alldistribT):
assert(len(weights)==alldistribT.shape[1])
return np.exp(np.dot(np.log(alldistribT),weights.T))
def geometricMean(alldistribT):
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, tol_error=1e-4,log=dict()):
"""Compute the Regularizzed wassersteien barycenter of distributions A"""
if weights is None:
weights=np.ones(A.shape[1])/A.shape[1]
else:
assert(len(weights)==A.shape[1])
#compute Mmax once for all
#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
log['niter']=0
log['all_err']=[]
while (err>tol_error 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['all_err'].append(err)
log['niter']=cpt
return geometricBar(weights,UKv),log
def unmix(distrib,D,M,M0,h0,reg,reg0,alpha,numItermax = 1000, tol_error=1e-3,log=dict()):
"""
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])
"""
#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':[]}
log['niter']=0
log['all_err']=[]
while (err>tol_error and cpt<numItermax):
K = projC(K,distrib)
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
log['all_err'].append(err)
cpt = cpt+1
log['niter']=cpt
return np.sum(K0,axis=1),log
|
