我要反转某个位置的地理编码纬度和经度,并希望获取该位置的地址。我已经通过Google Web Service完成了,但是需要时间。我想知道是否还有其他好的有效方法。
当前正在调用此服务,
NSString * getAddress = [NSString stringWithFormat:@"http://maps.googleapis.com/maps/api/geocode/json?latlng=%@,%@&sensor=true",Lattitude,Longitude];
您可以使用CLGeocoder
:
[self.geocoder reverseGeocodeLocation:location completionHandler:^(NSArray *placemarks, NSError *error){
CLPlacemark *placemark = placemarks[0];
NSLog(@"Found %@", placemark.name);
}];
尽管这仍然需要时间,因为这两种方法都使用Web服务将经/纬度转换为位置
尝试此代码。
geoCoder = [[CLGeocoder alloc]init];
[self.geoCoder reverseGeocodeLocation: locationManager.location completionHandler:
//Getting Human readable Address from Lat long,,,
^(NSArray *placemarks, NSError *error) {
//Get nearby address
CLPlacemark *placemark = [placemarks objectAtIndex:0];
//String to hold address
NSString *locatedAt = [[placemark.addressDictionary valueForKey:@"FormattedAddressLines"] componentsJoinedByString:@", "];
//Print the location to console
NSLog(@"I am currently at %@",locatedAt);
}];
看看GLGeocoder
。具体来说reverseGeocodeLocation:completionHandler:
您可以使用Apple的CLGeocoder
(CoreLocation的一部分)。具体来说,– reverseGeocodeLocation:completionHandler:
方法将返回给定坐标的地址数据字典。
看看this tutorial,或者,如果只是想快速复制一些内容:NSArray * addressOutput;CLLocation * currentLocation;//假定这些实例变量
// Reverse Geocoding
CLGeocoder *geocoder = [[CLGeocoder alloc] init];
[geocoder reverseGeocodeLocation:currentLocation completionHandler:^(NSArray *placemarks, NSError *error) {
NSLog(@"Found placemarks: %@, error: %@", placemarks, error);
if (error == nil && [placemarks count] > 0) {
NSMutableArray *tempArray = [[NSMutableArray alloc] initWithCapacity:[placemarks count]];
for (CLPlacemark *placemark in placemarks) {
[tempArray addObject:[NSString stringWithFormat:@"%@ %@\n%@ %@\n%@\n%@",
placemark.subThoroughfare, placemark.thoroughfare,
placemark.postalCode, placemark.locality,
placemark.administrativeArea,
placemark.country]];
}
addressOutput = [tempArray copy];
}
else {
addressOutput = nil;
NSLog(@"%@", error.debugDescription);
}
}];
基于本教程中的代码。
如果您不想使用Google API,请尝试以下代码-基本上将纬度和经度输入转换为ZIP(可以调整为地址)。
pip install uszipcode
# Import packages
from uszipcode import SearchEngine
search = SearchEngine(simple_zipcode=True)
from uszipcode import Zipcode
import numpy as np
#define zipcode search function
def get_zipcode(lat, lon):
result = search.by_coordinates(lat = lat, lng = lon, returns = 1)
return result[0].zipcode
#load columns from dataframe
lat = df_shooting['Latitude']
lon = df_shooting['Longitude']
#define latitude/longitude for function
df = pd.DataFrame({'lat':lat, 'lon':lon})
#add new column with generated zip-code
df['zipcode'] = df.apply(lambda x: get_zipcode(x.lat,x.lon), axis=1)
#print result
print(df)
#(optional) save as csv
#df.to_csv(r'zip_codes.csv')