Re: [cctbxbb] iterating over crystals

17 Sep 2023

      Thanks everyone for the great suggestions!

I think there are multiple ways I could have got this to work, but in the
end, as I suspected, cctbx *does* already have a ready-made function. Us
human crystallographers might not have been able to find it immediately,
but some assistance from ChatGPT could unearth it. All credit to Peter
Zwart for acting as "prompt engineer" here. It wasn't totally
straightforward, as shown by the conversation at the link below. I think
it's a good example of how to use these tools though. One has to be
insistent in the face of confabulation. From Peter's email:

https://chat.openai.com/share/b0a3ca0b-01ab-4469-a965-b54b4a3fed3d
...
which essentially tells me to look at the source code
Computational Crystallography Toolbox / Code / [r25775]
/trunk/cctbx/development/random_structure.py (sourceforge.net)
https://sourceforge.net/p/cctbx/code/HEAD/tree/trunk/cctbx/development/rando...
go to line 344, it uses
space_group_info.any_compatible_unit_cell(volume=1000)
This is what my loop looks like now, following these pointers:

for space_group_symbol in sgtbx.space_group_symbol_iterator():
    space_group = sgtbx.space_group(space_group_symbol)
    unit_cell = space_group.info().any_compatible_unit_cell(volume=1000)
    print(space_group_symbol.universal_hermann_mauguin(),
unit_cell.parameters())

Admittedly the cells do not explore the full random space within the
constraints, but actually for my purposes that's fine.

For what it's worth, this was Bard's attempt:
https://g.co/bard/share/7340bcdde4b3

This sort of looks promising, but fails because space_group.bravais_lattice
does not exist. I got disillusioned at that point, but what Peter's
conversation shows is that you've just got to keep pushing. At least,
that's the case right now. I wonder how long before these tools just write
the whole test I wanted correctly the first time.

Cheers
-- David

On Sat, 16 Sept 2023 at 09:33, Aaron Brewster  wrote:
...
Hi David, there is a constraints object that tells you which elements of
the metrical matrix can be varied:
...
...
...
from cctbx.sgtbx import space_group_info
sg = space_group_info('P4').group()
c = sg.adp_constraints()
c.independent_indices
(1, 2)
This means elements 1 and 2 of the metrical matrix are independent and can
be varied.  So you could randomly vary those elements of the
metrical matrix.  Say here we pick (25,100).  Then to convert those back to
a unit cell:
...
...
...
from cctbx.uctbx import unit_cell
uc =
unit_cell(metrical_matrix=c.all_params(independent_params=(25,100)))
print(uc)
(5, 5, 10, 90, 90, 90)
This code is part of symmetrize_reduce_enlarge in
./rstbx/symmetry/constraints/parameter_reduction.py
I would presume this works for any space group.  Is this along the lines
you were looking for?
-Aaron
On Fri, Sep 15, 2023 at 11:59 AM David Waterman 
wrote:
...
Hi folks,
I'm testing some code changes and would like to iterate over every space
group and for each generate a random unit cell compatible with the Bravais
symmetry. I've got the iteration sorted with space_group_symbol_iterator,
but I'm not sure how best to generate some compatible unit cell from the
space_group object. Is this something I could do easily with cctbx?
Cheers,
-- David
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