If you want to compute all possible Miller indices to a given resolution, for a unit cell, without considering symmetry something like this is what you need:
def generate_reflection_indices(uc, dmin):
maxh, maxk, maxl = uc.max_miller_indices(dmin)
indices = []
for h in range(-maxh, maxh + 1):
for k in range(-maxk, maxk + 1):
for l in range(-maxl, maxl + 1):
if h == 0 and k == 0 and l == 0:
continue
if uc.d((h, k, l)) < dmin:
continue
indices.append((h, k, l))
return indices
This does however mean that systematically absent reflections will be included - you would need to also test this with the space group object
Cheers Graeme
On 6 Dec 2017, at 00:48, Haiguang Liu mailto:[email protected]> wrote:
Hi Peter,
Will the command ms.expand_to_P1() do the trick to generate all symmetry peaks?
Best,
Haiguang
On Wed, Dec 6, 2017 at 2:13 AM, Peter Zwart mailto:[email protected]> wrote:
Hi Xuanxuan,
This:
from cctbx import miller
import cctbx
from cctbx import crystal
ms = miller.build_set(
crystal_symmetry=crystal.symmetry(
space_group_symbol='Fd-3m',
unit_cell=(5.4307,5.4307,5.4307,90.00,90.0,90.00)),
anomalous_flag=False,
d_min=0.4)
for hkl in ms.indices():
print hkl
gives:
(0, 2, 2)
.
.
.
(1, 3, 3)
etc
2,2,0 is related by symmetry to 0,2,2
HTH-P
On 5 December 2017 at 01:01, 李选选 mailto:[email protected]> wrote:
Dear cctbx developers,
I’m Xuanxuan, and I’m using cctbx to develop a crystallography-related algorithm. Recently, I found there is something wrong when I use miller package to generate miller indices for given space group and unit cell parameters. Maybe it’s my mistake, or it’s a bug.
Here are some details. I just use the following code to generate a list of miller indices for a Fd-3m crystal.
ms = miller.build_set(
crystal_symmetry=crystal.symmetry(
space_group_symbol='Fd-3m',
unit_cell=(5.4307,5.4307,5.4307,90.00,90.0,90.00)),
anomalous_flag=False,
d_min=0.4)
And I found some reflections are missing, like 220, 311, which should exist according to the reflection conditions listed in this page http://img.chem.ucl.ac.uk/sgp/large/227az2.htm.
Looking forward to your response.
Thanks,
Xuanxuan
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