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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_structures(self, s_list, anonymous=False):
 """ Given a list of structures, use fit to group them by structural equality. Args: s_list ([Structure]):
 List of structures to be grouped anonymous (bool):
 Wheher to use anonymous mode. Returns: A list of lists of matched structures Assumption: if s1 == s2 but s1 != s3, than s2 and s3 will be put in different groups without comparison. """

        if self._subset:
            raise ValueError("allow_subset cannot be used with"
                             " group_structures")

        original_s_list = list(s_list)
        s_list = self._process_species(s_list)

        # Use structure hash to pre-group structures
        if anonymous:
            c_hash = lambda c: c.anonymized_formula
        else:
            c_hash = self._comparator.get_hash
        s_hash = lambda s: c_hash(s[1].composition)
        sorted_s_list = sorted(enumerate(s_list), key=s_hash)
        all_groups = []

        # For each pre-grouped list of structures, perform actual matching.
        for k, g in itertools.groupby(sorted_s_list, key=s_hash):
            unmatched = list(g)
            while len(unmatched) > 0:
                i, refs = unmatched.pop(0)
                matches = [i]
                if anonymous:
                    inds = filter(lambda i: self.fit_anonymous(refs,
                            unmatched[i][1]), list(range(len(unmatched))))
                else:
                    inds = filter(lambda i: self.fit(refs, unmatched[i][1]),
                                  list(range(len(unmatched))))
                inds = list(inds)
                matches.extend([unmatched[i][0] for i in inds])
                unmatched = [unmatched[i] for i in range(len(unmatched))
                             if i not in inds]
                all_groups.append([original_s_list[i] for i in matches])

        return all_groups