On Sep 7, 2018, at 22:22, Alexey Orlov <data2sc...@gmail.com> wrote: > I'm trying to calculate the number of equivalent/nonequivalent neighbor > heteroatoms for each atom i of molecule m. > > For examples, the third carbon atom of molecule CC(OH)CC has two > nonequivalent neighbors: one carbon atom connected to OH and CH3 and another > carbon atom that is connected only to H. In the case of molecule CC(OH)C(C)C > the third carbon atom has two equivalent neighbor heteroatoms (methyl groups) > and another neighbor distinct from these two -- atom connected to OH and CH3.
I think you are looking for Chem.CanonicalRankAtoms() described at http://rdkit.org/Python_Docs/rdkit.Chem.rdmolfiles-module.html#CanonicalRankAtoms . >>> mol = Chem.MolFromSmiles("CC(O)CC") >>> list(Chem.CanonicalRankAtoms(mol, breakTies=False)) [1, 4, 2, 3, 0] As you wrote, there are no equivalent atoms. >>> mol = Chem.MolFromSmiles("CC(O)C(C)C") >>> list(Chem.CanonicalRankAtoms(mol, breakTies=False)) [2, 5, 3, 4, 0, 0] this identifies the two identical methyl groups, both with rank 0. You might try something like: from collections import Counter def get_atoms_neighbor_rank_counts(mol): indices = [] ranks = list(Chem.CanonicalRankAtoms(mol, breakTies=False)) for atom in mol.GetAtoms(): equivalents = Counter() equivalents.update(ranks[a.GetIdx()] for a in atom.GetNeighbors()) counts = [count for rank, count in equivalents.most_common()] print("Neighbor symmetry counts for", atom.GetIdx(), "are", counts) For your first test case it gives: >>> mol = Chem.MolFromSmiles("CC(O)CC") >>> get_atoms_neighbor_rank_counts(mol) Neighbor symmetry counts for 0 are [1] Neighbor symmetry counts for 1 are [1, 1, 1] Neighbor symmetry counts for 2 are [1] Neighbor symmetry counts for 3 are [1, 1] Neighbor symmetry counts for 4 are [1] That is, all of the atoms have unique neighbors. For the second test case it gives: >>> mol = Chem.MolFromSmiles("CC(O)C(C)C") >>> get_atoms_neighbor_rank_counts(mol) Neighbor symmetry counts for 0 are [1] Neighbor symmetry counts for 1 are [1, 1, 1] Neighbor symmetry counts for 2 are [1] Neighbor symmetry counts for 3 are [2, 1] Neighbor symmetry counts for 4 are [1] Neighbor symmetry counts for 5 are [1] That is, the atom with index 3 (the 4th atom) has 3 neighbors, two of which are identical. Finally, >>> mol = Chem.MolFromSmiles("CC(C)(C)OC(C)(C)C") >>> get_atoms_neighbor_rank_counts(mol) Neighbor symmetry counts for 0 are [1] Neighbor symmetry counts for 1 are [3, 1] Neighbor symmetry counts for 2 are [1] Neighbor symmetry counts for 3 are [1] Neighbor symmetry counts for 4 are [2] Neighbor symmetry counts for 5 are [3, 1] Neighbor symmetry counts for 6 are [1] Neighbor symmetry counts for 7 are [1] Neighbor symmetry counts for 8 are [1] This says that the atoms with indices 1 and 5 each have 4 neighbor, 3 of which are identical, and that the atom with index 4 has 2 neighbors, both of which are identical. Andrew da...@dalkescientific.com _______________________________________________ Rdkit-discuss mailing list Rdkit-discuss@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/rdkit-discuss
