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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def group_molecules(self, mol_list):
 """ Group molecules by structural equality. Args: mol_list: List of OpenBabel OBMol or pymatgen objects Returns: A list of lists of matched molecules Assumption: if s1=s2 and s2=s3, then s1=s3 This may not be true for small tolerances. """

        mol_hash = [(i, self._mapper.get_molecule_hash(m))
                    for i, m in enumerate(mol_list)]
        mol_hash.sort(key=lambda x: x[1])

        #Use molecular hash to pre-group molecules.
        raw_groups = tuple([tuple([m[0] for m in g]) for k, g
                            in itertools.groupby(mol_hash,
                                                 key=lambda x: x[1])])

        group_indices = []
        for rg in raw_groups:
            mol_eq_test = [(p[0], p[1], self.fit(mol_list[p[0]],
                                                 mol_list[p[1]]))
                           for p in itertools.combinations(sorted(rg), 2)]
            mol_eq = set([(p[0], p[1]) for p in mol_eq_test if p[2]])
            not_alone_mols = set(itertools.chain.from_iterable(mol_eq))
            alone_mols = set(rg) - not_alone_mols
            group_indices.extend([[m] for m in alone_mols])
            while len(not_alone_mols) > 0:
                current_group = {not_alone_mols.pop()}
                while len(not_alone_mols) > 0:
                    candidate_pairs = set(
                        [tuple(sorted(p)) for p
                         in itertools.product(current_group, not_alone_mols)])
                    mutual_pairs = candidate_pairs & mol_eq
                    if len(mutual_pairs) == 0:
                        break
                    mutual_mols = set(itertools.chain
                                      .from_iterable(mutual_pairs))
                    current_group |= mutual_mols
                    not_alone_mols -= mutual_mols
                group_indices.append(sorted(current_group))

        group_indices.sort(key=lambda x: (len(x), -x[0]), reverse=True)
        all_groups = [[mol_list[i] for i in g] for g in group_indices]
        return all_groups