Geographical queries made easy.
Project description
pygeons
Geographical queries made simple.
Free software: MIT license
Documentation: 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']
Features
Determine if a (city, state and country) combination corresponds to an existing place name
Scrub (city, state, country) combinations
Normalize city, state and country names to their canonical representations
Frame queries in English as well as languages native to each particular country
Credits
This package was created with Cookiecutter and the audreyr/cookiecutter-pypackage project template.
History
0.1.1 (2017-12-03)
First working release. Includes import scripts and source code.
0.1.0 (2017-11-26)
First release on PyPI.
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Source Distribution
pygeons-0.1.1.tar.gz
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