from __future__ import print_function
# Copyright (C) 2010, Jesper Friis
# (see accompanying license files for details).
"""Utility tools for convenient creation of slabs and interfaces of
different orientations."""
import numpy as np
[docs]def wrap_positions(positions, cell, pbc=True, center=(0.5, 0.5, 0.5),
eps=1e-7):
"""Wrap positions to unit cell.
Returns positions changed by a multiple of the unit cell vectors to
fit inside the space spanned by these vectors. See also the
:meth:`ase.atoms.Atoms.wrap` method.
Parameters:
positions: float ndarray of shape (n, 3)
Positions of the atoms
cell: float ndarray of shape (3, 3)
Unit cell vectors.
pbc: one or 3 bool
For each axis in the unit cell decides whether the positions
will be moved along this axis.
center: three float
The positons in fractional coordinates that the new positions
will be nearest possible to.
eps: float
Small number to prevent slightly negative coordinates from beeing
wrapped.
Example:
>>> from ase.utils.geometry import wrap_positions
>>> wrap_positions([[-0.1, 1.01, -0.5]],
... [[1, 0, 0], [0, 1, 0], [0, 0, 4]],
... pbc=[1, 1, 0])
array([[ 0.9 , 0.01, -0.5 ]])
"""
if not hasattr(pbc, '__len__'):
pbc = (pbc,) * 3
shift = np.asarray(center) - 0.5 + eps * np.asarray(pbc, dtype=bool)
fractional = np.linalg.solve(np.asarray(cell).T,
np.asarray(positions).T).T + shift
for i, periodic in enumerate(pbc):
if periodic:
fractional[:, i] %= 1.0
fractional[:, i] -= shift[i]
return np.dot(fractional, cell)
[docs]def get_layers(atoms, miller, tolerance=0.001):
"""Returns two arrays describing which layer each atom belongs
to and the distance between the layers and origo.
Parameters:
miller: 3 integers
The Miller indices of the planes. Actually, any direction
in reciprocal space works, so if a and b are two float
vectors spanning an atomic plane, you can get all layers
parallel to this with miller=np.cross(a,b).
tolerance: float
The maximum distance in Angstrom along the plane normal for
counting two atoms as belonging to the same plane.
Returns:
tags: array of integres
Array of layer indices for each atom.
levels: array of floats
Array of distances in Angstrom from each layer to origo.
Example:
>>> import numpy as np
>>> from ase.lattice.spacegroup import crystal
>>> atoms = crystal('Al', [(0,0,0)], spacegroup=225, cellpar=4.05)
>>> np.round(atoms.positions, decimals=5)
array([[ 0. , 0. , 0. ],
[ 0. , 2.025, 2.025],
[ 2.025, 0. , 2.025],
[ 2.025, 2.025, 0. ]])
>>> get_layers(atoms, (0,0,1))
(array([0, 1, 1, 0]), array([ 0. , 2.025]))
"""
miller = np.asarray(miller)
metric = np.dot(atoms.cell, atoms.cell.T)
c = np.linalg.solve(metric.T, miller.T).T
miller_norm = np.sqrt(np.dot(c, miller))
d = np.dot(atoms.get_scaled_positions(), miller) / miller_norm
keys = np.argsort(d)
ikeys = np.argsort(keys)
mask = np.concatenate(([True], np.diff(d[keys]) > tolerance))
tags = np.cumsum(mask)[ikeys]
if tags.min() == 1:
tags -= 1
levels = d[keys][mask]
return tags, levels
[docs]def cut(atoms, a=(1, 0, 0), b=(0, 1, 0), c=None, clength=None,
origo=(0, 0, 0), nlayers=None, extend=1.0, tolerance=0.01,
maxatoms=None):
"""Cuts out a cell defined by *a*, *b*, *c* and *origo* from a
sufficiently repeated copy of *atoms*.
Typically, this function is used to create slabs of different
sizes and orientations. The vectors *a*, *b* and *c* are in scaled
coordinates and defines the returned cell and should normally be
integer-valued in order to end up with a periodic
structure. However, for systems with sub-translations, like fcc,
integer multiples of 1/2 or 1/3 might also make sence for some
directions (and will be treated correctly).
Parameters:
atoms: Atoms instance
This should correspond to a repeatable unit cell.
a: int | 3 floats
The a-vector in scaled coordinates of the cell to cut out. If
integer, the a-vector will be the scaled vector from *origo* to the
atom with index *a*.
b: int | 3 floats
The b-vector in scaled coordinates of the cell to cut out. If
integer, the b-vector will be the scaled vector from *origo* to the
atom with index *b*.
c: None | int | 3 floats
The c-vector in scaled coordinates of the cell to cut out.
if integer, the c-vector will be the scaled vector from *origo* to
the atom with index *c*.
If *None* it will be along cross(a, b) converted to real space
and normalised with the cube root of the volume. Note that this
in general is not perpendicular to a and b for non-cubic
systems. For cubic systems however, this is redused to
c = cross(a, b).
clength: None | float
If not None, the length of the c-vector will be fixed to
*clength* Angstroms. Should not be used together with
*nlayers*.
origo: int | 3 floats
Position of origo of the new cell in scaled coordinates. If
integer, the position of the atom with index *origo* is used.
nlayers: None | int
If *nlayers* is not *None*, the returned cell will have
*nlayers* atomic layers in the c-direction.
extend: 1 or 3 floats
The *extend* argument scales the effective cell in which atoms
will be included. It must either be three floats or a single
float scaling all 3 directions. By setting to a value just
above one, e.g. 1.05, it is possible to all the corner and
edge atoms in the returned cell. This will of cause make the
returned cell non-repeatable, but is very usefull for
visualisation.
tolerance: float
Determines what is defined as a plane. All atoms within
*tolerance* Angstroms from a given plane will be considered to
belong to that plane.
maxatoms: None | int
This option is used to auto-tune *tolerance* when *nlayers* is
given for high zone axis systems. For high zone axis one
needs to reduce *tolerance* in order to distinguise the atomic
planes, resulting in the more atoms will be added and
eventually MemoryError. A too small *tolerance*, on the other
hand, might result in inproper splitting of atomic planes and
that too few layers are returned. If *maxatoms* is not None,
*tolerance* will automatically be gradually reduced until
*nlayers* atomic layers is obtained, when the number of atoms
exceeds *maxatoms*.
Example:
>>> import ase
>>> from ase.lattice.spacegroup import crystal
>>>
# Create an aluminium (111) slab with three layers
#
# First an unit cell of Al
>>> a = 4.05
>>> aluminium = crystal('Al', [(0,0,0)], spacegroup=225,
... cellpar=[a, a, a, 90, 90, 90])
>>>
# Then cut out the slab
>>> al111 = cut(aluminium, (1,-1,0), (0,1,-1), nlayers=3)
>>>
# Visualisation of the skutterudite unit cell
#
# Again, create a skutterudite unit cell
>>> a = 9.04
>>> skutterudite = crystal(
... ('Co', 'Sb'),
... basis=[(0.25,0.25,0.25), (0.0, 0.335, 0.158)],
... spacegroup=204,
... cellpar=[a, a, a, 90, 90, 90])
>>>
# Then use *origo* to put 'Co' at the corners and *extend* to
# include all corner and edge atoms.
>>> s = cut(skutterudite, origo=(0.25, 0.25, 0.25), extend=1.01)
>>> ase.view(s) # doctest: +SKIP
"""
atoms = atoms.copy()
cell = atoms.cell
if isinstance(origo, int):
origo = atoms.get_scaled_positions()[origo]
origo = np.array(origo, dtype=float)
scaled = (atoms.get_scaled_positions() - origo) % 1.0
scaled %= 1.0 # needed to ensure that all numbers are *less* than one
atoms.set_scaled_positions(scaled)
if isinstance(a, int):
a = scaled[a] - origo
if isinstance(b, int):
b = scaled[b] - origo
if isinstance(c, int):
c = scaled[c] - origo
a = np.array(a, dtype=float)
b = np.array(b, dtype=float)
if c is None:
metric = np.dot(cell, cell.T)
vol = np.sqrt(np.linalg.det(metric))
h = np.cross(a, b)
H = np.linalg.solve(metric.T, h.T)
c = vol * H / vol**(1. / 3.)
c = np.array(c, dtype=float)
if nlayers:
# Recursive increase the length of c until we have at least
# *nlayers* atomic layers parallell to the a-b plane
while True:
at = cut(atoms, a, b, c, origo=origo, extend=extend,
tolerance=tolerance)
scaled = at.get_scaled_positions()
d = scaled[:, 2]
keys = np.argsort(d)
ikeys = np.argsort(keys)
tol = tolerance
while True:
mask = np.concatenate(([True], np.diff(d[keys]) > tol))
tags = np.cumsum(mask)[ikeys] - 1
levels = d[keys][mask]
if (maxatoms is None or len(at) < maxatoms or
len(levels) > nlayers):
break
tol *= 0.9
if len(levels) > nlayers:
break
c *= 2
at.cell[2] *= levels[nlayers]
return at[tags < nlayers]
newcell = np.dot(np.array([a, b, c]), cell)
if nlayers is None and clength is not None:
newcell[2, :] *= clength / np.linalg.norm(newcell[2])
# Create a new atoms object, repeated and translated such that
# it completely covers the new cell
scorners_newcell = np.array([[0., 0., 0.], [0., 0., 1.],
[0., 1., 0.], [0., 1., 1.],
[1., 0., 0.], [1., 0., 1.],
[1., 1., 0.], [1., 1., 1.]])
corners = np.dot(scorners_newcell, newcell * extend)
scorners = np.linalg.solve(cell.T, corners.T).T
rep = np.ceil(scorners.ptp(axis=0)).astype('int') + 1
trans = np.dot(np.floor(scorners.min(axis=0)), cell)
atoms = atoms.repeat(rep)
atoms.translate(trans)
atoms.set_cell(newcell)
# Mask out atoms outside new cell
stol = 0.1 * tolerance # scaled tolerance, XXX
maskcell = atoms.cell * extend
sp = np.linalg.solve(maskcell.T, (atoms.positions).T).T
mask = np.all(np.logical_and(-stol <= sp, sp < 1 - stol), axis=1)
atoms = atoms[mask]
return atoms
[docs]class IncompatibleCellError(ValueError):
"""Exception raised if stacking fails due to incompatible cells
between *atoms1* and *atoms2*."""
pass
[docs]def stack(atoms1, atoms2, axis=2, cell=None, fix=0.5,
maxstrain=0.5, distance=None, reorder=False,
output_strained=False):
"""Return a new Atoms instance with *atoms2* stacked on top of
*atoms1* along the given axis. Periodicity in all directions is
ensured.
The size of the final cell is determined by *cell*, except
that the length alongh *axis* will be the sum of
*atoms1.cell[axis]* and *atoms2.cell[axis]*. If *cell* is None,
it will be interpolated between *atoms1* and *atoms2*, where
*fix* determines their relative weight. Hence, if *fix* equals
zero, the final cell will be determined purely from *atoms1* and
if *fix* equals one, it will be determined purely from
*atoms2*.
An ase.geometry.IncompatibleCellError exception is raised if the
cells of *atoms1* and *atoms2* are incopatible, e.g. if the far
corner of the unit cell of either *atoms1* or *atoms2* is
displaced more than *maxstrain*. Setting *maxstrain* to None,
disable this check.
If *distance* is not None, the size of the final cell, along the
direction perpendicular to the interface, will be adjusted such
that the distance between the closest atoms in *atoms1* and
*atoms2* will be equal to *distance*. This option uses
scipy.optimize.fmin() and hence require scipy to be installed.
If *reorder* is True, then the atoms will be reordred such that
all atoms with the same symbol will follow sequensially after each
other, eg: 'Al2MnAl10Fe' -> 'Al12FeMn'.
If *output_strained* is True, then the strained versions of
*atoms1* and *atoms2* are returned in addition to the stacked
structure.
Example:
>>> import ase
>>> from ase.lattice.spacegroup import crystal
>>>
# Create an Ag(110)-Si(110) interface with three atomic layers
# on each side.
>>> a_ag = 4.09
>>> ag = crystal(['Ag'], basis=[(0,0,0)], spacegroup=225,
... cellpar=[a_ag, a_ag, a_ag, 90., 90., 90.])
>>> ag110 = cut(ag, (0, 0, 3), (-1.5, 1.5, 0), nlayers=3)
>>>
>>> a_si = 5.43
>>> si = crystal(['Si'], basis=[(0,0,0)], spacegroup=227,
... cellpar=[a_si, a_si, a_si, 90., 90., 90.])
>>> si110 = cut(si, (0, 0, 2), (-1, 1, 0), nlayers=3)
>>>
>>> interface = stack(ag110, si110, maxstrain=1)
>>> ase.view(interface) # doctest: +SKIP
>>>
# Once more, this time adjusted such that the distance between
# the closest Ag and Si atoms will be 2.3 Angstrom (requires scipy).
>>> interface2 = stack(ag110, si110,
... maxstrain=1, distance=2.3) # doctest:+ELLIPSIS
Optimization terminated successfully.
...
>>> ase.view(interface2) # doctest: +SKIP
"""
atoms1 = atoms1.copy()
atoms2 = atoms2.copy()
if (np.sign(np.linalg.det(atoms1.cell)) !=
np.sign(np.linalg.det(atoms2.cell))):
raise IncompatibleCellError('*atoms1* amd *atoms2* must both either '
'have a lefthanded or a righanded cell.')
c1 = np.linalg.norm(atoms1.cell[axis])
c2 = np.linalg.norm(atoms2.cell[axis])
if cell is None:
cell1 = atoms1.cell.copy()
cell2 = atoms2.cell.copy()
cell1[axis] /= c1
cell2[axis] /= c2
cell = cell1 + fix * (cell2 - cell1)
cell[axis] /= np.linalg.norm(cell[axis])
cell1 = cell.copy()
cell2 = cell.copy()
cell1[axis] *= c1
cell2[axis] *= c2
if maxstrain:
strain1 = np.sqrt(((cell1 - atoms1.cell).sum(axis=0)**2).sum())
strain2 = np.sqrt(((cell2 - atoms2.cell).sum(axis=0)**2).sum())
if strain1 > maxstrain or strain2 > maxstrain:
raise IncompatibleCellError(
'*maxstrain* exceeded. *atoms1* strained %f and '
'*atoms2* strained %f.' % (strain1, strain2))
atoms1.set_cell(cell1, scale_atoms=True)
atoms2.set_cell(cell2, scale_atoms=True)
if output_strained:
atoms1_strained = atoms1.copy()
atoms2_strained = atoms2.copy()
if distance is not None:
from scipy.optimize import fmin
def mindist(pos1, pos2):
n1 = len(pos1)
n2 = len(pos2)
idx1 = np.arange(n1).repeat(n2)
idx2 = np.tile(np.arange(n2), n1)
return np.sqrt(((pos1[idx1] - pos2[idx2])**2).sum(axis=1).min())
def func(x):
t1, t2, h1, h2 = x[0:3], x[3:6], x[6], x[7]
pos1 = atoms1.positions + t1
pos2 = atoms2.positions + t2
d1 = mindist(pos1, pos2 + (h1 + 1.0) * atoms1.cell[axis])
d2 = mindist(pos2, pos1 + (h2 + 1.0) * atoms2.cell[axis])
return (d1 - distance)**2 + (d2 - distance)**2
atoms1.center()
atoms2.center()
x0 = np.zeros((8,))
x = fmin(func, x0)
t1, t2, h1, h2 = x[0:3], x[3:6], x[6], x[7]
atoms1.translate(t1)
atoms2.translate(t2)
atoms1.cell[axis] *= 1.0 + h1
atoms2.cell[axis] *= 1.0 + h2
atoms2.translate(atoms1.cell[axis])
atoms1.cell[axis] += atoms2.cell[axis]
atoms1.extend(atoms2)
if reorder:
atoms1 = sort(atoms1)
if output_strained:
return atoms1, atoms1_strained, atoms2_strained
else:
return atoms1
[docs]def sort(atoms, tags=None):
"""Return a new Atoms object with sorted atomic order. The default
is to order according to chemical symbols, but if *tags* is not
None, it will be used instead. A stable sorting algorithm is used.
Example:
>>> import ase
>>> from ase.lattice.spacegroup import crystal
>>>
# Two unit cells of NaCl
>>> a = 5.64
>>> nacl = crystal(['Na', 'Cl'], [(0, 0, 0), (0.5, 0.5, 0.5)],
... spacegroup=225, cellpar=[a, a, a, 90, 90, 90]).repeat((2, 1, 1))
>>> nacl.get_chemical_symbols()
['Na', 'Na', 'Na', 'Na', 'Cl', 'Cl', 'Cl', 'Cl', 'Na', 'Na', 'Na',
'Na', 'Cl', 'Cl', 'Cl', 'Cl']
>>> nacl_sorted = sort(nacl)
>>> nacl_sorted.get_chemical_symbols()
['Cl', 'Cl', 'Cl', 'Cl', 'Cl', 'Cl', 'Cl', 'Cl', 'Na', 'Na', 'Na',
'Na', 'Na', 'Na', 'Na', 'Na']
>>> np.all(nacl_sorted.cell == nacl.cell)
True
"""
if tags is None:
tags = atoms.get_chemical_symbols()
else:
tags = list(tags)
deco = sorted([(tag, i) for i, tag in enumerate(tags)])
indices = [i for tag, i in deco]
return atoms[indices]
[docs]def rotation_matrix(a1, a2, b1, b2):
"""Returns a rotation matrix that rotates the vectors *a1* in the
direction of *a2* and *b1* in the direction of *b2*.
In the case that the angle between *a2* and *b2* is not the same
as between *a1* and *b1*, a proper rotation matrix will anyway be
constructed by first rotate *b2* in the *b1*, *b2* plane.
"""
a1 = np.asarray(a1, dtype=float) / np.norm(a1)
b1 = np.asarray(b1, dtype=float) / np.norm(b1)
c1 = np.cross(a1, b1)
c1 /= np.norm(c1) # clean out rounding errors...
a2 = np.asarray(a2, dtype=float) / np.norm(a2)
b2 = np.asarray(b2, dtype=float) / np.norm(b2)
c2 = np.cross(a2, b2)
c2 /= np.norm(c2) # clean out rounding errors...
# Calculate rotated *b2*
theta = np.arccos(np.dot(a2, b2)) - np.arccos(np.dot(a1, b1))
b3 = np.sin(theta) * a2 + np.cos(theta) * b2
b3 /= np.norm(b3) # clean out rounding errors...
A1 = np.array([a1, b1, c1])
A2 = np.array([a2, b3, c2])
R = np.linalg.solve(A1, A2).T
return R
[docs]def rotate(atoms, a1, a2, b1, b2, rotate_cell=True, center=(0, 0, 0)):
"""Rotate *atoms*, such that *a1* will be rotated in the direction
of *a2* and *b1* in the direction of *b2*. The point at *center*
is fixed. Use *center='COM'* to fix the center of mass. If
*rotate_cell* is true, the cell will be rotated together with the
atoms.
Note that the 000-corner of the cell is by definition fixed at
origo. Hence, setting *center* to something other than (0, 0, 0)
will rotate the atoms out of the cell, even if *rotate_cell* is
True.
"""
if isinstance(center, str) and center.lower() == 'com':
center = atoms.get_center_of_mass()
R = rotation_matrix(a1, a2, b1, b2)
atoms.positions[:] = np.dot(atoms.positions - center, R.T) + center
if rotate_cell:
atoms.cell[:] = np.dot(atoms.cell, R.T)
[docs]def minimize_tilt_ij(atoms, modified=1, fixed=0, fold_atoms=True):
"""Minimize the tilt angle for two given axes.
The problem is underdetermined. Therefore one can choose one axis
that is kept fixed.
"""
orgcell_cc = atoms.get_cell()
pbc_c = atoms.get_pbc()
i = fixed
j = modified
if not (pbc_c[i] and pbc_c[j]):
raise RuntimeError('Axes have to be periodic')
prod_cc = np.dot(orgcell_cc, orgcell_cc.T)
cell_cc = 1. * orgcell_cc
nji = np.floor(- prod_cc[i, j] / prod_cc[i, i] + 0.5)
cell_cc[j] = orgcell_cc[j] + nji * cell_cc[i]
# sanity check
def volume(cell):
return np.abs(np.dot(cell[2], np.cross(cell[0], cell[1])))
V = volume(cell_cc)
assert(abs(volume(orgcell_cc) - V) / V < 1.e-10)
atoms.set_cell(cell_cc)
if fold_atoms:
atoms.set_scaled_positions(atoms.get_scaled_positions())
[docs]def minimize_tilt(atoms, order=range(3), fold_atoms=True):
"""Minimize the tilt angles of the unit cell."""
pbc_c = atoms.get_pbc()
for i1, c1 in enumerate(order):
for c2 in order[i1 + 1:]:
if pbc_c[c1] and pbc_c[c2]:
minimize_tilt_ij(atoms, c1, c2, fold_atoms)
[docs]def find_mic(D, cell, pbc=True):
"""Finds the minimum-image representation of vector(s) D"""
# Calculate the 4 unique unit cell diagonal lengths
diags = np.sqrt((np.dot([[1, 1, 1],
[-1, 1, 1],
[1, -1, 1],
[-1, -1, 1],
], cell)**2).sum(1))
# calculate 'mic' vectors (D) and lengths (D_len) using simple method
Dr = np.dot(D, np.linalg.inv(cell))
D = np.dot(Dr - np.round(Dr) * pbc, cell)
D_len = np.sqrt((D**2).sum(1))
# return mic vectors and lengths for only orthorhombic cells,
# as the results may be wrong for non-orthorhombic cells
if (max(diags) - min(diags)) / max(diags) < 1e-9:
return D, D_len
# The cutoff radius is the longest direct distance between atoms
# or half the longest lattice vector, whichever is smaller
cutoff = min(max(D_len), max(diags) / 2.)
# The number of neighboring images to search in each direction is
# equal to the ceiling of the cutoff distance (defined above) divided
# by the length of the projection of the lattice vector onto its
# corresponding surface normal. a's surface normal vector is e.g.
# b x c / (|b| |c|), so this projection is (a . (b x c)) / (|b| |c|).
# The numerator is just the lattice volume, so this can be simplified
# to V / (|b| |c|). This is rewritten as V |a| / (|a| |b| |c|)
# for vectorization purposes.
latt_len = np.sqrt((cell**2).sum(1))
V = abs(np.linalg.det(cell))
n = pbc * np.array(np.ceil(cutoff * np.prod(latt_len) /
(V * latt_len)), dtype=int)
# Construct a list of translation vectors. For example, if we are
# searching only the nearest images (27 total), tvecs will be a
# 27x3 array of translation vectors. This is the only nested loop
# in the routine, and it takes a very small fraction of the total
# execution time, so it is not worth optimizing further.
tvecs = []
for i in range(-n[0], n[0] + 1):
latt_a = i * cell[0]
for j in range(-n[1], n[1] + 1):
latt_ab = latt_a + j * cell[1]
for k in range(-n[2], n[2] + 1):
tvecs.append(latt_ab + k * cell[2])
tvecs = np.array(tvecs)
# Translate the direct displacement vectors by each translation
# vector, and calculate the corresponding lengths.
D_trans = tvecs[np.newaxis] + D[:, np.newaxis]
D_trans_len = np.sqrt((D_trans**2).sum(2))
# Find mic distances and corresponding vector(s) for each given pair
# of atoms. For symmetrical systems, there may be more than one
# translation vector corresponding to the MIC distance; this finds the
# first one in D_trans_len.
D_min_len = np.min(D_trans_len, axis=1)
D_min_ind = D_trans_len.argmin(axis=1)
D_min = D_trans[range(len(D_min_ind)), D_min_ind]
return D_min, D_min_len
# Self test
if __name__ == '__main__':
import doctest
print('doctest: ', doctest.testmod())