pygeons 0.1.1

侏儒

地理查询变得简单。

• 自由软件：麻省理工学院许可证
• 文档：https://pygeons.readthedocs.io。

>>> from pprint import pprint
>>> import pygeons
>>> # Scrub a (city, state, country) combination
>>> scrubbed = pygeons.csc_scrub('sydney', 'nsw', 'au')
>>> result = scrubbed.pop('result')
>>> scrubbed
{'score': 0.9, 'st_status': 'O', 'cc_status': 'O', 'count': 1}
>>> pprint(result)
{'_id': 2147714,
 'abbr': [],
 'admin1': 'State of New South Wales',
 'admin1names': ['new south wales', 'nsw', 'state of new south wales'],
 'admin2': '17200',
 'admin2names': [],
 'asciiname': 'Sydney',
 'countryCode': 'AU',
 'featureClass': 'P',
 'featureCode': 'PPLA',
 'latitude': -33.86785,
 'longitude': 151.20732,
 'name': 'Sydney',
 'names': ['syd', 'sydney', 'sydney city'],
 'names_lang': {'en': ['syd', 'sydney', 'sydney city']},
 'population': 4627345}
>>> pygeons.csc_scrub('sydney', 'nsw', None)['result']['_id']  # same GeoNames ID as above
2147714
>>> # Normalize a state abbreviation
>>> pygeons.norm('admin1', 'AU', 'nsw')
'State of New South Wales'
>>> # Translate a country name in the native language into English
>>> pygeons.country_info('россия')['names_lang']['en'][:4]
['ru', 'rus', 'russia', 'russian federation']

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