I think you probably used a slightly different SMILES than the one you
showed. The one you showed should have given ((0,1,3,4),(2,1,3,4)).
The proper merge rule would then be to consider all matches equivalent if
the 2nd and 3rd atom in the match agree, in any order; i.e, the two
carbons, indices 1 and 3 in this case.
So to do this, for each molecule, do something like this:
d = dict{}
for match in matches:
t = (match[1], match[2])
if match[1] < match[2] ):
t = (match[1], match[2])
else:
t = (match[2], match[1])
d[t] = match
You will wind up with as many dictionary elements as there are matches.
-P.
On Tue, Nov 7, 2017 at 7:38 PM, James T. Metz via Rdkit-discuss <
rdkit-discuss@lists.sourceforge.net> wrote:
> RDkit Discussion Group,
>
> I have written a SMARTS to detect vicinal chlorine groups
> using RDkit. There are 4 atoms involved in a vicinal chlorine group.
>
> SMARTS = '[Cl]-[C,c]-,=,:[C,c]-[Cl]'
>
> I am trying to count the number of ("unique") occurrences of this
> pattern.
>
> For some molecules with symmetry, this results in
> over-counting.
>
> For the molecule, smiles1 below, I want to obtain
> a count of 1 i.e., 1 tuple of 4 atoms.
>
> smiles1 = 'ClC(Cl)CCl'
>
> However, using the SMARTS above, I obtain 2 tuples of 4 atoms.
> Beginning with a MOL file representation of smiles1, I get
>
> ((1,2,4,3), (0,2,4,3))
>
> One possible solution is to somehow merge the two tuples according
> to a "rule." One rule that works is "if 3 of the atom indices are the
> same,
> then combine into one tuple."
>
> However, the rule needs a bit of modification for more complicated
> cases (higher symmetry).
>
> Consider
>
> smiles2 = 'ClC(Cl)CCl(Cl)(Cl)
>
> My goal is to get 2 tuples of 4 atoms for smiles2
>
> smiles2 is somewhat tricky because there are either
> 2 groups of 3 (4 atom) tuples, or 3 groups of 2 (4 atom)
> tuples depending on how you choose your 3 atom indices.
>
> Again, if my goal is to get 2 tuples, then I need to somehow
> pick the largest group, i.e., 2 groups of 3 tuples to do the merge
> operation which will give me 2 remaining groups (desired).
>
> I have already checked stackoverflow and a few other places
> for PYTHON code to do the necessary merging, but I could not
> find anything specific and appropriate.
>
> I would be most grateful if anyone has ideas how to do this. I
> suspect the answer is a few lines of well-written PYTHON code,
> and not modifying the SMARTS (I could be mistaken!).
>
> Thank you.
>
> Regards,
> Jim Metz
>
>
>
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