轻量级python库按邮政编码查询城市/州名称？

Question

这里很简单,我正在寻找一个轻量级的库,它允许我查找给定邮政编码的城市/州配对.我正在使用django FWIW.提前致谢.

Answer 1

试试pyzipcode.主页上的一个例子:

>>> from pyzipcode import ZipCodeDatabase
>>> zcdb = ZipCodeDatabase()
>>> zipcode = zcdb[54115]
>>> zipcode.zip
u'54115'
>>> zipcode.city
u'De Pere'
>>> zipcode.state
u'WI'
>>> zipcode.longitude
-88.078959999999995
>>> zipcode.latitude
44.42042
>>> zipcode.timezone
-6

Answer 2

使用此库uszipcode.

好处:

• 数据是最新的最新的,超级富豪信息,方式更丰富,更先进的日期比zipcode和pyzipcode任何其他Python邮编库.
• 查询非常简单,您可以使用20多种内置查询模式.您可以随意自定义查询.
• 支持城市和州的模糊字符串匹配.您不需要使用确切的名称.

>>> from uszipcode import ZipcodeSearchEngine
>>> search = ZipcodeSearchEngine()
>>> zipcode = search.by_zipcode("10001")
>>> print(zipcode)
{
    "City": "New York", 
    "Density": 34035.48387096774, 
    "HouseOfUnits": 12476, 
    "LandArea": 0.62, 
    "Latitude": 40.75368539999999, 
    "Longitude": -73.9991637, 
    "NEBoundLatitude": 40.8282129, 
    "NEBoundLongitude": -73.9321059, 
    "Population": 21102, 
    "SWBoundLatitude": 40.743451, 
    "SWBoungLongitude": -74.00794499999998, 
    "State": "NY", 
    "TotalWages": 1031960117.0, 
    "WaterArea": 0.0, 
    "Wealthy": 48903.42702113544, 
    "Zipcode": "10001", 
    "ZipcodeType": "Standard"
}

# fuzzy city, state search, case insensitive, spelling mistake tolerant
# all zipcode in new york
>>> result = search.by_city_and_state(city="newyork", state="NY")
>>> search.export_to_csv(result, "result.csv")

非常容易构建高级搜索

>>> result = search.find(city="new york", 
... wealthy=100000, sort_by="Wealthy", ascending=False, returns=10)