地理距离得到城市蟒蛇

Question

我不确定这是否可行，但我有一个城市坐标列表，然后我想取列表中的第一个坐标并计算列表中每个坐标的距离，最后我想得到最短的距离和城市名称。除了获取城市名称和州之外，我能够做所有的事情。反正有吗？

from geopy.geocoders import GoogleV3
from geopy.distance import vincenty
geolocator = GoogleV3()

cord_places = [(41.6592776, -76.7503693), (37.7792768, -122.4192704), (42.3486635, -83.0567375)
first = cord_places[0];
cord_places.remove(first)

print distance(first, cord_places)
def distance(first, cord_places):
    for ct in cord_paces
        dist = vincenty((first), (ct))#this gives me all the distances between first city and the others in the list but I would like to know which city it was
    return dist

有人有任何提示吗？

Answer 1

如果您希望/需要使用Nominatim检查GoogleV3属性并查看google api doc .raw地理编码器会更简单>

from geopy.geocoders import Nominatim
geolocator = Nominatim()

def city_and_state(coord):
    location = geolocator.reverse(coord, exactly_one=True)
    address = location.raw['address']
    city = address.get('city', '')
    state = address.get('state', '')
    return city, state

如果地址未与城市或州相关联，则返回空字符串。对于您的第一个坐标(41.6592776, -76.7503693)，您可能需要hamlet密钥而不是city。