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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 exhaustive_ontology_ilx_diff_row_only( self, ontology_row: dict ) -> dict:
        ''' WARNING RUNTIME IS AWEFUL '''
        results = []
        header = ['Index'] + list(self.existing_ids.columns)
        for row in self.existing_ids.itertuples():
            row = {header[i]:val for i, val in enumerate(row)}
            check_list = [
                {
                    'external_ontology_row': ontology_row,
                    'ilx_rows': [row],
                },
            ]
            # First layer for each external row. Second is for each potential ilx row. It's simple here 1-1.
            result = self.__exhaustive_diff(check_list)[0][0]
            if result['same']:
                results.append(result)
        return results