ios: What is the efficient and recommended way to reverse geocode? - iphone

Question

I am to reverse Geo Code Latitude and Longitude of some location and want to get address of that location. I have done it through google web service but it takes time.
I want to know if there is some other good and efficient approach.
Currently calling this service,
NSString * getAddress = [NSString stringWithFormat:#"http://maps.googleapis.com/maps/api/geocode/json?latlng=%#,%#&sensor=true",Lattitude,Longitude];

Answer 1

You can use CLGeocoder:
[self.geocoder reverseGeocodeLocation:location completionHandler:^(NSArray *placemarks, NSError *error){
CLPlacemark *placemark = placemarks[0];
NSLog(#"Found %#", placemark.name);
}];
This will still take time though, since both methods use web services to convert lat / long into a place

Answer 2

Try This code .
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);
}];

Answer 3

Have a look at GLGeocoder. Specifically reverseGeocodeLocation:completionHandler:

Answer 4

You can use Apple's CLGeocoder (part of CoreLocation). Specifically, the – reverseGeocodeLocation:completionHandler: method which will return a dictionary of address data for the given coordinates.

Answer 5

Have a look at this tutorial, or, if just want something to copy quickly: NSArray *addressOutput;
CLLocation *currentLocation;
//assumes these instance variables
// 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);
}
}];
Based off the code in the tutorial.

Answer 6

If you to not want to use google API, try this code - basically transforms latitude and longitude inputs into ZIPs (can be adjusted to adresses).
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')