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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 _feature_most_alternative(self, results, full_results=False):
 """ Find the placename with the most alternative names and return its country. More alternative names are a rough measure of importance. Paramaters results: dict output of `query_geonames` Returns ------- most_alt: str ISO code of country of place with most alternative names, or empty string if none """

        try:
            alt_names = [len(i['alternativenames']) for i in results['hits']['hits']]
            most_alt = results['hits']['hits'][np.array(alt_names).argmax()]
            if full_results:
                return most_alt
            else:
                return most_alt['country_code3']
        except (IndexError, ValueError, TypeError):
            return ""