def structk(strucA, strucB, alchem=alchemy(), periodic=False, mode="match", fout=None, peps=0.0, gamma=1.0, zeta=1.0, xspecies=False):
# computes the SOAP similarity KERNEL between two structures by combining atom-centered kernels
# possible kernel modes include:
# average : scalar product between averaged kernels
# match: best-match hungarian kernel
# permanent: average over all permutations
# average kernel. quick & easy!
if mode=="fastavg":
genvA=strucA.globenv
genvB=strucB.globenv
return envk(genvA, genvB, alchem)**zeta, 0
elif mode=="fastspecies":
# for now, only implement standard Kronecker alchemy
senvB = environ(strucB.nmax, strucB.lmax, strucB.alchem)
kk = 0
for za in strucA.zspecies:
if not za in strucB.zspecies: continue
senvA = environ(strucA.nmax, strucA.lmax, strucA.alchem)
for ia in xrange(strucA.getnz(za)):
senvA.add(strucA.getenv(za, ia))
senvB = environ(strucB.nmax, strucB.lmax, strucB.alchem)
for ib in xrange(strucB.getnz(za)):
senvB.add(strucB.getenv(za, ib))
kk += envk(senvA, senvB, alchem)**zeta
kk/=strucA.nenv*strucB.nenv
return kk,0
# for zb, nzb in nspeciesB:
# for ib in xrange(nzb):
# return envk(genvA, genvB, alchem), 0
nenv = 0
if periodic: # replicate structures to match structures of different periodicity
# we do not check for compatibility at this stage, just assume that the
# matching will be done somehow (otherwise it would be exceedingly hard to manage in case of non-standard alchemy)
nspeciesA = []
nspeciesB = []
for z in strucA.zspecies:
nspeciesA.append( (z, strucA.getnz(z)) )
for z in strucB.zspecies:
nspeciesB.append( (z, strucB.getnz(z)) )
nenv=nenvA = strucA.nenv
nenvB = strucB.nenv
else:
# top up missing atoms with isolated environments
# first checks which atoms are present
zspecies = sorted(list(set(strucB.zspecies+strucA.zspecies)))
nspecies = []
for z in zspecies:
nz = max(strucA.getnz(z),strucB.getnz(z))
nspecies.append((z,nz))
nenv += nz
nenvA = nenvB = nenv
nspeciesA = nspeciesB = nspecies
np.set_printoptions(linewidth=500,precision=4)
kk = np.zeros((nenvA,nenvB),float)
ika = 0
ikb = 0
for za, nza in nspeciesA:
for ia in xrange(nza):
envA = strucA.getenv(za, ia)
ikb = 0
for zb, nzb in nspeciesB:
for ib in xrange(nzb):
envB = strucB.getenv(zb, ib)
if alchem.mu > 0 and (strucA.ismissing(za, ia) ^ strucB.ismissing(zb, ib)):
# includes a penalty dependent on "mu", in a way that is consistent with the definition of kernel distance
kk[ika,ikb] = exp(-alchem.mu)
else:
if za == zb or not xspecies: #uncomment to zero out kernels between different species
kk[ika,ikb] = envk(envA, envB, alchem)**zeta
else: kk[ika,ikb] = 0
ikb+=1
ika+=1
aidx = {}
ika=0
for za, nza in nspeciesA:
aidx[za] = range(ika,ika+nza)
ika+=nza
ikb=0
bidx = {}
for zb, nzb in nspeciesB:
bidx[zb] = range(ikb,ikb+nzb)
ikb+=nzb
if fout != None:
# prints out similarity information for the environment pairs
fout.write("# atomic species in the molecules (possibly topped up with dummy isolated atoms): \n")
for za, nza in nspeciesA:
for ia in xrange(nza): fout.write(" %d " % (za) )
fout.write("\n");
for zb, nzb in nspeciesB:
for ib in xrange(nzb): fout.write(" %d " % (zb) )
fout.write("\n");
fout.write("# environment kernel matrix: \n")
for r in kk:
for e in r:
fout.write("%20.14e " % (e) )
fout.write("\n")
#fout.write("# environment kernel eigenvalues: \n")
#ev = np.linalg.eigvals(kk)
#for e in ev:
# fout.write("(%8.4e,%8.4e) " % (e.real,e.imag) )
#fout.write("\n");
# Now we have the matrix of scalar products.
# We can first find the optimal scalar product kernel
# we must find the maximum "cost"
if mode == "match":
if periodic and nenvA != nenvB:
nenv = lcm(nenvA, nenvB)
hun = lcm_best_cost(1-kk)
else:
hun=best_cost(1.0-kk)
cost = 1-hun/nenv
elif mode == "permanent":
# there is no place to hide: cross-species environments are not necessarily zero
if peps>0: cost = mcperm(kk, peps)
else: cost = xperm(kk)
cost = cost/np.math.factorial(nenv)/nenv
elif mode == "rematch":
cost=rematch(kk, gamma, 1e-6) # hard-coded residual error for regularized gamma
# print cost, kk.sum()/(nenv*nenv), envk(strucA.globenv, strucB.globenv, alchem)
elif mode == "average":
cost = kk.sum()/(nenvA*nenvB)
# print 'elem: {}'.format(kk.sum())
# print 'elem norm: {}'.format(cost)
# print 'avg norm: {}'.format((nenvA*nenvB))
else: raise ValueError("Unknown global fingerprint mode ", mode)
return cost,kk
def structk(strucA,
strucB,
alchem=alchemy(),
peratom=False,
mode="match",
fout=None,
peps=0.0,
gamma=1.0,
zeta=1.0,
xspecies=False):
# computes the SOAP similarity KERNEL between two structures by combining atom-centered kernels
# possible kernel modes include:
# average : scalar product between averaged kernels
# match: best-match hungarian kernel
# permanent: average over all permutations
# average kernel. quick & easy!
if mode == "fastavg":
genvA = strucA.globenv
genvB = strucB.globenv
return envk(genvA, genvB, alchem)**zeta, 0
elif mode == "fastspecies":
# for now, only implement standard Kronecker alchemy
senvB = environ(strucB.nmax, strucB.lmax, strucB.alchem)
kk = 0
for za in strucA.zspecies:
if not za in strucB.zspecies: continue
senvA = environ(strucA.nmax, strucA.lmax, strucA.alchem)
for ia in xrange(strucA.getnz(za)):
senvA.add(strucA.getenv(za, ia))
senvB = environ(strucB.nmax, strucB.lmax, strucB.alchem)
for ib in xrange(strucB.getnz(za)):
senvB.add(strucB.getenv(za, ib))
kk += envk(senvA, senvB, alchem)**zeta
kk /= strucA.nenv * strucB.nenv
return kk, 0
# for zb, nzb in nspeciesB:
# for ib in xrange(nzb):
# return envk(genvA, genvB, alchem), 0
nenv = 0
if peratom: # replicate structures to match structures of different peratomity
# we do not check for compatibility at this stage, just assume that the
# matching will be done somehow (otherwise it would be exceedingly hard to manage in case of non-standard alchemy)
nspeciesA = []
nspeciesB = []
for z in strucA.zspecies:
nspeciesA.append((z, strucA.getnz(z)))
for z in strucB.zspecies:
nspeciesB.append((z, strucB.getnz(z)))
nenv = nenvA = strucA.nenv
nenvB = strucB.nenv
else:
# top up missing atoms with isolated environments
# first checks which atoms are present
zspecies = sorted(list(set(strucB.zspecies + strucA.zspecies)))
nspecies = []
for z in zspecies:
nz = max(strucA.getnz(z), strucB.getnz(z))
nspecies.append((z, nz))
nenv += nz
nenvA = nenvB = nenv
nspeciesA = nspeciesB = nspecies
np.set_printoptions(linewidth=500, precision=4)
kk = np.zeros((nenvA, nenvB), float)
ika = 0
ikb = 0
for za, nza in nspeciesA:
for ia in xrange(nza):
envA = strucA.getenv(za, ia)
ikb = 0
for zb, nzb in nspeciesB:
for ib in xrange(nzb):
envB = strucB.getenv(zb, ib)
if alchem.mu > 0 and (strucA.ismissing(za, ia)
^ strucB.ismissing(zb, ib)):
# includes a penalty dependent on "mu", in a way that is consistent with the definition of kernel distance
kk[ika, ikb] = exp(-alchem.mu)
else:
if za == zb or not xspecies: #uncomment to zero out kernels between different species
kk[ika, ikb] = envk(envA, envB, alchem)**zeta
else:
kk[ika, ikb] = 0
ikb += 1
ika += 1
aidx = {}
ika = 0
for za, nza in nspeciesA:
aidx[za] = range(ika, ika + nza)
ika += nza
ikb = 0
bidx = {}
for zb, nzb in nspeciesB:
bidx[zb] = range(ikb, ikb + nzb)
ikb += nzb
if fout != None:
# prints out similarity information for the environment pairs
fout.write(
"# atomic species in the molecules (possibly topped up with dummy isolated atoms): \n"
)
for za, nza in nspeciesA:
for ia in xrange(nza):
fout.write(" %d " % (za))
fout.write("\n")
for zb, nzb in nspeciesB:
for ib in xrange(nzb):
fout.write(" %d " % (zb))
fout.write("\n")
fout.write("# environment kernel matrix: \n")
for r in kk:
for e in r:
fout.write("%20.14e " % (e))
fout.write("\n")
#fout.write("# environment kernel eigenvalues: \n")
#ev = np.linalg.eigvals(kk)
#for e in ev:
# fout.write("(%8.4e,%8.4e) " % (e.real,e.imag) )
#fout.write("\n");
# Now we have the matrix of scalar products.
# We can first find the optimal scalar product kernel
# we must find the maximum "cost"
if mode == "match":
if peratom and nenvA != nenvB:
nenv = lcm(nenvA, nenvB)
hun = lcm_best_cost(1 - kk)
else:
hun = best_cost(1.0 - kk)
cost = 1 - hun / nenv
elif mode == "permanent":
# there is no place to hide: cross-species environments are not necessarily zero
if peps > 0: cost = mcperm(kk, peps)
else: cost = xperm(kk)
cost = cost / np.math.factorial(nenv) / nenv
elif mode == "rematch":
cost = rematch(kk, gamma,
1e-6) # hard-coded residual error for regularized gamma
# print cost, kk.sum()/(nenv*nenv), envk(strucA.globenv, strucB.globenv, alchem)
elif mode == "average":
cost = kk.sum() / (nenvA * nenvB)
# print 'elem: {}'.format(kk.sum())
# print 'elem norm: {}'.format(cost)
# print 'avg norm: {}'.format((nenvA*nenvB))
else:
raise ValueError("Unknown global fingerprint mode ", mode)
return cost, kk
def structk(strucA,
strucB,
alchem=alchemy(),
periodic=False,
mode="match",
fout=None,
peps=0.0,
gamma=1.0,
zeta=1.0,
xspecies=False):
# computes the SOAP similarity KERNEL between two structures by combining atom-centered kernels
# possible kernel modes include:
# average : scalar product between averaged kernels
# match: best-match hungarian kernel
# permanent: average over all permutations
# top up missing atoms with isolated environments
# first checks which atoms are present
zspecies = sorted(list(set(strucB.zspecies + strucA.zspecies)))
nspecies = []
nenv = 0
for z in zspecies:
nz = max(strucA.getnz(z), strucB.getnz(z))
nspecies.append((z, nz))
nenv += nz
# Make a mapping from atom idx to (species, species_idx)
idx_to_spec = [zip([z] * nz, range(nz)) for z, nz in nspecies]
idx_to_spec = [itm for sublist in idx_to_spec for itm in sublist]
nsp = len(zspecies)
# alchemical matrix for species
alchemAB = np.empty((nsp, nsp), dtype=float)
for sA in xrange(nsp):
for sB in xrange(sA + 1):
alchemAB[sA, sB] = alchem.getpair(zspecies[sB], zspecies[sA])
alchemAB[sB, sA] = alchemAB[sA, sB]
#prepares the lists of pairs to avoid calling many times getpair further down the line
#for i1 in xrange(nsp):
# s1 = zspecies[i1]
# for i2 in xrange(nsp):
# s2 = zspecies[i2]
# a=envA.getpair(s1,s2)
# b=envB.getpair(s1,s2)
# if i1==0 and i2==0:
# arrA = np.empty((nenv, nsp, nsp, a.shape[0]), float)
# arrB = np.empty((nenv, nsp, nsp, a.shape[0]), float)
# arrA[i1,i2,:] = a
# arrB[i1,i2,:] = b
#first = True
#for za, nza in nspecies:
# for ia in xrange(nza):
# envA = strucA.getenv(za, ia)
# ikb = 0
# for zb, nzb in nspecies:
# a = envA.getpair(za, zb)
# for ib in xrange(nzb):
# envB = strucB.getenv(zb, ib)
# b = envB.getpair(za, zb)
# if first:
# arrA = np.empty((nenv, nsp, nsp, a.shape[0]), float)
# arrB = np.empty((nenv, nsp, nsp, a.shape[0]), float)
# first = False
# arrB[ikb,i1,i2,:] = b
# ikb+=1
# arrA[ika, i1,i2,:] = a
# ika+=1
for ik in range(nenv):
i = idx_to_spec[ik]
envA = strucA.getenv(i[0], i[1])
envB = strucB.getenv(i[0], i[1])
for i1 in range(nsp):
s1 = zspecies[i1]
for i2 in range(nsp):
s2 = zspecies[i2]
a = envA.getpair(s1, s2)
b = envB.getpair(s1, s2)
if i1 == 0 and i2 == 0 and ik == 0:
arrA = np.empty((nenv, nsp, nsp, a.shape[0]), dtype=float)
arrB = np.empty((nenv, nsp, nsp, a.shape[0]), dtype=float)
arrA[ik, i1, i2, :] = a
arrB[ik, i1, i2, :] = b
kk = tensordot(arrA, arrB, alchemAB)
#dotprod = np.tensordot(arrA, arrB, axes=([3],[3]))
#dotprod = np.tensordot(dotprod, alchemAB, axes=([1,4],[0,1]))
#kk = np.tensordot(dotprod, alchemAB, axes=([1,3],[0,1]))
#kk = np.zeros((nenv,nenv),float)
#for ika in range(nenv):
# ai = idx_to_spec[ika]
# envA = strucA.getenv(ai[0], ai[1])
# for ikb in range(nenv):
# bi = idx_to_spec[ikb]
# envB = strucB.getenv(bi[0], bi[1])
# res = envk(envA, envB, alchem)**zeta
# kk[ika,ikb] = res
#ika = 0
#ikb = 0
#for za, nza in nspecies:
#for ia in xrange(nza):
#envA = strucA.getenv(za, ia)
#ikb = 0
#for zb, nzb in nspecies:
#for ib in xrange(nzb):
#envB = strucB.getenv(zb, ib)
#res = envk(envA, envB, alchem)**zeta
#kk[ika,ikb] = res
#ikb+=1
#ika+=1
# Now we have the matrix of scalar products.
# We can first find the optimal scalar product kernel
# we must find the maximum "cost"
if mode == "rematch":
cost = rematch(kk, gamma,
1e-6) # hard-coded residual error for regularized gamma
# print cost, kk.sum()/(nenv*nenv), envk(strucA.globenv, strucB.globenv, alchem)
else:
raise ValueError("Unknown global fingerprint mode ", mode)
return cost, kk