Hi Greg,
Thank you for your reply!
I tested the method you suggested, but it seems like
rdMolDraw2D.PrepareMolForDrawing() has no affect.
e.g, I have a molecule m, to which I add hydrogens. I then compute the 2D
coordinates of m, and use rdMolDraw2D.PrepareMolForDrawing() as you
suggested.
To test it, I compare the outputs of Chem.MolToMolBlock() on the molecule
and on the prepared drawing.
The results are the same, and it is defently not what you would expect from
a standard representation (i.e. the ketone oxygen is not vertically up from
its carbon, the chain is not oriented right, etc.).
See the following explicit example:
In[5]: m = Chem.MolFromSmiles('CCC(O)=O')
In[6]: m = Chem.AddHs(m)
In[7]: AllChem.Compute2DCoords(m)
In[8]: m_draw = Chem.Draw.rdMolDraw2D.PrepareMolForDrawing(m)
In[9]: print (Chem.MolToMolBlock(m))
RDKit 2D
11 10 0 0 0 0 0 0 0 0999 V2000
1.3490 -0.4340 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0
0.3421 0.6778 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0
-1.1242 0.3617 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0
-1.5836 -1.0662 0.0000 O 0 0 0 0 0 0 0 0 0 0 0 0
-2.1311 1.4735 0.0000 O 0 0 0 0 0 0 0 0 0 0 0 0
2.8153 -0.1180 0.0000 H 0 0 0 0 0 0 0 0 0 0 0 0
1.9149 -1.8232 0.0000 H 0 0 0 0 0 0 0 0 0 0 0 0
0.0792 -1.2326 0.0000 H 0 0 0 0 0 0 0 0 0 0 0 0
1.6119 1.4763 0.0000 H 0 0 0 0 0 0 0 0 0 0 0 0
-0.2238 2.0669 0.0000 H 0 0 0 0 0 0 0 0 0 0 0 0
-3.0499 -1.3823 0.0000 H 0 0 0 0 0 0 0 0 0 0 0 0
1 2 1 0
2 3 1 0
3 4 1 0
3 5 2 0
1 6 1 0
1 7 1 0
1 8 1 0
2 9 1 0
2 10 1 0
4 11 1 0
M END
In[10]: print (Chem.MolToMolBlock(m_draw))
RDKit 2D
11 10 0 0 0 0 0 0 0 0999 V2000
1.3490 -0.4340 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0
0.3421 0.6778 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0
-1.1242 0.3617 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0
-1.5836 -1.0662 0.0000 O 0 0 0 0 0 0 0 0 0 0 0 0
-2.1311 1.4735 0.0000 O 0 0 0 0 0 0 0 0 0 0 0 0
2.8153 -0.1180 0.0000 H 0 0 0 0 0 0 0 0 0 0 0 0
1.9149 -1.8232 0.0000 H 0 0 0 0 0 0 0 0 0 0 0 0
0.0792 -1.2326 0.0000 H 0 0 0 0 0 0 0 0 0 0 0 0
1.6119 1.4763 0.0000 H 0 0 0 0 0 0 0 0 0 0 0 0
-0.2238 2.0669 0.0000 H 0 0 0 0 0 0 0 0 0 0 0 0
-3.0499 -1.3823 0.0000 H 0 0 0 0 0 0 0 0 0 0 0 0
1 2 1 0
2 3 1 0
3 4 1 0
3 5 2 0
1 6 1 0
1 7 1 0
1 8 1 0
2 9 1 0
2 10 1 0
4 11 1 0
M END
Am I missing something?
Best,
Peleg
On Sun, Aug 13, 2017 at 11:59 AM, Greg Landrum <greg.land...@gmail.com>
wrote:
> Hi Peleg,
>
> On Sat, Aug 12, 2017 at 4:08 PM, Peleg Bar-Sapir <pel...@gmail.com> wrote:
>
>>
>> I'm trying to get the 2D coordinates and info of all the atoms and bonds,
>> including hydrogens (i.e. atom type, bond type, etc.) of a molecule in a
>> standard structural formula. To clarify: I'm not referring to
>> Chem.MolToMolBlock(), but the coordinates used by the Draw class (e.g. in
>> MolToFile()) to create standard representations.
>>
>> Any advice on how to extract those? Maybe one can reach them via one of
>> the Draw member functions/variables?
>>
>
> The drawing code calls rdMolDraw2D.PrepareMolForDrawing() before actually
> doing the drawing. This function, among other things, assigns coordinates.
> Assuming that you're looking for the coordinates in the molecule's
> coordinate space, you can get access to them by calling
> PrepareMolForDrawing() and then either getting a mol block:
>
> In [7]: m = Chem.AddHs(Chem.MolFromSmiles('CC#C'))
>
> In [8]: nm = rdMolDraw2D.PrepareMolForDrawing(m)
>
> In [9]: print(Chem.MolToMolBlock(nm))
>
> RDKit 2D
>
> 7 6 0 0 0 0 0 0 0 0999 V2000
> -0.6887 -0.0795 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0
> 0.8112 -0.0614 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0
> 2.3111 -0.0433 0.0000 C 0 0 0 0 0 0 0 0 0 0 0 0
> -1.4230 -1.3875 0.0000 H 0 0 0 0 0 0 0 0 0 0 0 0
> -2.1044 0.4165 0.0000 H 0 0 0 0 0 0 0 0 0 0 0 0
> -0.4461 1.4007 0.0000 H 0 0 0 0 0 0 0 0 0 0 0 0
> 1.5400 -0.2455 0.0000 H 0 0 0 0 0 0 0 0 0 0 0 0
> 1 2 1 0
> 2 3 3 0
> 1 4 1 0
> 1 5 1 0
> 1 6 1 0
> 3 7 1 0
> M END
>
>
> Or by using the molecule's conformer object:
>
> In [13]: conf = nm.GetConformer()
>
> In [14]: for aidx in range(conf.GetNumAtoms()):
> ...: pos = conf.GetAtomPosition(aidx)
> ...: print(pos.x, pos.y, pos.z)
> ...:
> -0.6887274130576998 -0.07950997820751476 0.0
> 0.8111632161215289 -0.0613964335423347 0.0
> 2.311053845300757 -0.04328288887715446 0.0
> -1.4229859378146839 -1.387510138307542 0.0
> -2.104358766425875 0.416461665499047 0.0
> -0.4461124978218231 1.4007393261007453 0.0
> 1.5399675536977953 -0.24550155266524612 0.0
>
>
> I'm not sure what kind of additional info you are looking for, but maybe
> this much helps?
>
> -greg
>
>
>
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