Hi Nik,

Thanks for your speedy reply.

On 21/01/13 17:19, Stiefl, Nikolaus wrote:
> do you just have to check if an atom is in a 6-membered ring? If so then
>
>
> In [8]: m =
> Chem.MolFromSmiles('COc1ccc(cc1O[C@H]1C[C@@H]2CC[C@H]1C2)C1CNC(=O)NC1')
>
> In [9]: [a.IsInRingSize(6) for a in m.GetAtoms()]
>
>
> Should help.

Hmmm! I thought I tried that!  I seem to get a different result:

    RDKit::RingInfo* ring_info_p = rdkm.getRingInfo();
    for (unsigned int iat=0; iat<rdkm.getNumAtoms(); iat++) {
       bool b = ring_info_p->isAtomInRingOfSize(iat, 6);
       std::string name;
       rdkm[iat]->getProp("name", name);
       std::cout << iat << "  " << name << " " << b << std::endl;
    }

results in:

0   O3   0
1   C17  1
2   N1   1
3   HN1  0
4   N2   1
5   HN2  0
6   C18  1
7   H18  0
8  H18A  0
9   C15  1
10   H15  0
11   C16  1
12  H16A  0
13   H16  0
14   C12  1
15   C13  1
16   H13  0
17   C14  1
18   H14  0
19   C2   1
20   O1   0
21   C1   0
22   H1B  0
23   H1A  0
24   H1   0
25   C11  1
26   H11  0
27   C3   1
28   O2   0
29   C4   0
30   H4   0
31   C9   0
32   H9   0
33   C8   0
34   H8   0
35   H8A  0
36   C7   0
37   H7A  0
38   H7   0
39   C10  0
40   H10  0
41  H10A  0
42   C6   0
43   H6   0
44   C5   0
45   H5A  0
46   H5   0

The ring atoms of interest are C4, C5, C6, C7, C8, C9, (C10 is the 
bridging carbon)

Anyway, this will fix norbornane - but not the general case (I'm 
thinking of amantadine and the like - see below).

Oh, it just occurred to me that I didn't start with the SMILES string (I 
created an rdkit molecule from the restraints).  I will try the SMILES 
string.

>
> Sorry - maybe I do not fully understand your question
>

It was poorly formed, sorry. I see now that there is another part I 
neglected to mention.  I need to know all the 3,4,5,6-membered rings 
that an atom might be part of (actually, to be strictly accurate, I need 
to know the number of each of such rings)  so the atoms of norbornane 
would result in 5,5 (one atom) or 5,6 (four such atoms) or 5,5,6 (two 
such atoms).

Thanks,

Paul.


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