If you're just looking for 6 membered rings, you can define a SMARTS that
matches 6 membered rings like this "*1~*~*~*~*~*1". You can also use this
approach to identify all rings (at least those within reason). You can use
an expression like this
["*1"+string.join(["*~"]*x,"")+"*1" for x in range(1,19)]
to generate SMARTS for all rings with size 3 to 20. Now you can match
these to your molecule and get all of the rings (example below).
import string
from rdkit import Chem
class RingFinder:
def __init__(self):
self.ringSmartsList = ["*1"+string.join(["*~"]*x,"")+"*1" for x in
range(1,19)]
self.ringPatList = [(x.count("*"),Chem.MolFromSmarts(x)) for x in
self.ringSmartsList]
def findAllRings(self,mol):
ringList = []
for size,pat in self.ringPatList:
for match in mol.GetSubstructMatches(pat):
ringList.append([size,match])
return ringList
ringFinder = RingFinder()
smiles = "COc1ccc(cc1O[C@H]1C[C@@H]2CC[C@H]1C2)C1CNC(=O)NC1"
mol = Chem.MolFromSmiles(smiles)
print ringFinder.findAllRings(mol)
If you run this you'll get two 5 membered rings and 3 six membered rings
for the molecule above.
Pat
On Mon, Jan 21, 2013 at 10:13 AM, Paul Emsley <pems...@mrc-lmb.cam.ac.uk>wrote:
>
> I am making heavy weather of the following problem - and am wondering if I
> am missing something (such as a useful RDKit function).
>
> I am working on this beasty (as an example):
>
> http://www.rcsb.org/pdb/**ligand/ligandsummary.do?hetId=**0CP<http://www.rcsb.org/pdb/ligand/ligandsummary.do?hetId=0CP>
>
> COc1ccc(cc1O[C@H]1C[C@@H]2CC[**C@H]1C2)C1CNC(=O)NC1
>
> which has a norbornane substituent. I am trying to prepare input for a
> downstream program that needs to know if the norbornane atoms are in a
> 6-membered ring [1]. RingInfo gives me the 2 5-membered rings. I am
> strugging to make use of that information to find 6-membered rings. I have
> been using makeRingNeighborMap() and pickFusedRings(). Am I missing an
> RDKit function that finds all rings?
>
> Cheers,
>
> Paul.
>
>
> [1] actually, all atoms but it is the norbornane atoms with which I
> struggle.
>
>
>
>
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