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# -*- coding: utf-8 -*-
"""
==============================
1D Wasserstein barycenter demo
==============================
@author: rflamary
"""
import numpy as np
import matplotlib.pylab as pl
import ot
from mpl_toolkits.mplot3d import Axes3D #necessary for 3d plot even if not used
import scipy as sp
import scipy.signal as sps
#%% parameters
n=10 # nb bins
# bin positions
x=np.arange(n,dtype=np.float64)
xx,yy=np.meshgrid(x,x)
xpos=np.hstack((xx.reshape(-1,1),yy.reshape(-1,1)))
M=ot.dist(xpos)
I0=((xx-5)**2+(yy-5)**2<3**2)*1.0
I1=((xx-7)**2+(yy-7)**2<3**2)*1.0
I0/=I0.sum()
I1/=I1.sum()
i0=I0.ravel()
i1=I1.ravel()
M=M[i0>0,:][:,i1>0].copy()
i0=i0[i0>0]
i1=i1[i1>0]
Itot=np.concatenate((I0[:,:,np.newaxis],I1[:,:,np.newaxis]),2)
#%% plot the distributions
pl.figure(1)
pl.subplot(2,2,1)
pl.imshow(I0)
pl.subplot(2,2,2)
pl.imshow(I1)
#%% barycenter computation
alpha=0.5 # 0<=alpha<=1
weights=np.array([1-alpha,alpha])
def conv2(I,k):
return sp.ndimage.convolve1d(sp.ndimage.convolve1d(I,k,axis=1),k,axis=0)
def conv2n(I,k):
res=np.zeros_like(I)
for i in range(I.shape[2]):
res[:,:,i]=conv2(I[:,:,i],k)
return res
def get_1Dkernel(reg,thr=1e-16,wmax=1024):
w=max(min(wmax,2*int((-np.log(thr)*reg)**(.5))),3)
x=np.arange(w,dtype=np.float64)
return np.exp(-((x-w/2)**2)/reg)
thr=1e-16
reg=1e0
k=get_1Dkernel(reg)
pl.figure(2)
pl.plot(k)
I05=conv2(I0,k)
pl.figure(1)
pl.subplot(2,2,1)
pl.imshow(I0)
pl.subplot(2,2,2)
pl.imshow(I05)
#%%
G=ot.emd(i0,i1,M)
r0=np.sum(M*G)
reg=1e-1
Gs=ot.bregman.sinkhorn_knopp(i0,i1,M,reg=reg)
rs=np.sum(M*Gs)
#%%
def mylog(u):
tmp=np.log(u)
tmp[np.isnan(tmp)]=0
return tmp
def sinkhorn_conv(a,b, reg, numItermax = 1000, stopThr=1e-9, verbose=False, log=False,**kwargs):
a=np.asarray(a,dtype=np.float64)
b=np.asarray(b,dtype=np.float64)
if len(b.shape)>2:
nbb=b.shape[2]
a=a[:,:,np.newaxis]
else:
nbb=0
if log:
log={'err':[]}
# we assume that no distances are null except those of the diagonal of distances
if nbb:
u = np.ones((a.shape[0],a.shape[1],nbb))/(np.prod(a.shape[:2]))
v = np.ones((a.shape[0],a.shape[1],nbb))/(np.prod(b.shape[:2]))
a0=1.0/(np.prod(b.shape[:2]))
else:
u = np.ones((a.shape[0],a.shape[1]))/(np.prod(a.shape[:2]))
v = np.ones((a.shape[0],a.shape[1]))/(np.prod(b.shape[:2]))
a0=1.0/(np.prod(b.shape[:2]))
k=get_1Dkernel(reg)
if nbb:
K=lambda I: conv2n(I,k)
else:
K=lambda I: conv2(I,k)
cpt = 0
err=1
while (err>stopThr and cpt<numItermax):
uprev = u
vprev = v
v = np.divide(b, K(u))
u = np.divide(a, K(v))
if (np.any(np.isnan(u)) or np.any(np.isnan(v))
or np.any(np.isinf(u)) or np.any(np.isinf(v))):
# we have reached the machine precision
# come back to previous solution and quit loop
print('Warning: numerical errors at iteration', cpt)
u = uprev
v = vprev
break
if cpt%10==0:
# we can speed up the process by checking for the error only all the 10th iterations
err = np.sum((u-uprev)**2)/np.sum((u)**2)+np.sum((v-vprev)**2)/np.sum((v)**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
if log:
log['u']=u
log['v']=v
if nbb: #return only loss
res=np.zeros((nbb))
for i in range(nbb):
res[i]=np.sum(u[:,i].reshape((-1,1))*K*v[:,i].reshape((1,-1))*M)
if log:
return res,log
else:
return res
else: # return OT matrix
res=reg*a0*np.sum(a*mylog(u+(u==0))+b*mylog(v+(v==0)))
if log:
return res,log
else:
return res
reg=1e0
r,log=sinkhorn_conv(I0,I1,reg,verbose=True,log=True)
a=I0
b=I1
u=log['u']
v=log['v']
#%% barycenter interpolation
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