我正在使用GPS座標排行榜的iPhone應用程序。我不需要精確的座標---實際上我並不想的座標是準確的，以保護用戶的隱私。有沒有簡單的方法讓GPS座標變得粗糙？
我指定的kCLLocationAccuracyThreeKilometers爲desiredAccuracy，但是當GPS是活動的，似乎它也可以拿起設備的確切位置。
問題：有沒有簡單的算法可以用來使GPS數據更粗？說，使它粒度爲3公里。
如果我只是按比例縮小數字並刪除小數點並再次縮小它們，它會使它在世界某些地方比其他地方更粗糙。
謝謝！
儘管Mark的回答很有用，但仍然沒有產生一致結果的公式，因爲它依賴於隨機數生成器。
我的好友爲此提供了最好的回答：
回合的緯度，經度取決於粒度最近顯著的身影，但是這將導致所有附近的某個位置的緯度/經度，風中相同的位置。該方法將使用緯度/經度中兩點之間的距離來計算緯度的四捨五入。使用下面的公式，並將路線設置爲0，那麼距離就是您的距離粒度。計算產生的新緯度/經度減去兩個緯度/經度以獲得緯度的舍入量。然後將標題設置爲90並重新計算並從舊中減去新的經緯度以獲得lon的舍入量。
而這裏的C++代碼：
class LocationUtility
{
public: static Location getLocationNow()
{
Location location;
if(context != null)
{
double latitude = 0;
double longitude = 0;
::of_getCurrentLocation(&latitude, &longitude);
location.setLatitude(latitude);
location.setLongitude(longitude);
location = makeLocationCoarse(location);
}
return location;
}
public: static Location makeLocationCoarse(const Location& location)
{
double granularityInMeters = 3 * 1000;
return makeLocationCoarse(location, granularityInMeters);
}
public: static Location makeLocationCoarse(const Location& location,
double granularityInMeters)
{
Location courseLocation;
if(location.getLatitude() == (double)0 &&
location.getLongitude() == (double)0)
{
// Special marker, don't bother.
}
else
{
double granularityLat = 0;
double granularityLon = 0;
{
// Calculate granularityLat
{
double angleUpInRadians = 0;
Location newLocationUp = getLocationOffsetBy(location,
granularityInMeters, angleUpInRadians);
granularityLat = location.getLatitude() -
newLocationUp.getLatitude();
if(granularityLat < (double)0)
{
granularityLat = -granularityLat;
}
}
// Calculate granularityLon
{
double angleRightInRadians = 1.57079633;
Location newLocationRight = getLocationOffsetBy(location,
granularityInMeters, angleRightInRadians);
granularityLon = location.getLongitude() -
newLocationRight.getLongitude();
if(granularityLon < (double)0)
{
granularityLon = -granularityLon;
}
}
}
double courseLatitude = location.getLatitude();
double courseLongitude = location.getLongitude();
{
if(granularityLon == (double)0 || granularityLat == (double)0)
{
courseLatitude = 0;
courseLongitude = 0;
}
else
{
courseLatitude = (int)(courseLatitude/granularityLat) *
granularityLat;
courseLongitude = (int)(courseLongitude/granularityLon) *
granularityLon;
}
}
courseLocation.setLatitude(courseLatitude);
courseLocation.setLongitude(courseLongitude);
}
return courseLocation;
}
// http://www.movable-type.co.uk/scripts/latlong.html
private: static Location getLocationOffsetBy(const Location& location,
double offsetInMeters, double angleInRadians)
{
Location newLocation;
double lat1 = location.getLatitude();
double lon1 = location.getLongitude();
lat1 = deg2rad(lat1);
lon1 = deg2rad(lon1);
double distanceKm = offsetInMeters/(double)1000;
const double earthRadiusKm = 6371;
double lat2 = asin(sin(lat1)*cos(distanceKm/earthRadiusKm) +
cos(lat1)*sin(distanceKm/earthRadiusKm)*cos(angleInRadians));
double lon2 = lon1 +
atan2(sin(angleInRadians)*sin(distanceKm/earthRadiusKm)*cos(lat1),
cos(distanceKm/earthRadiusKm)-sin(lat1)*sin(lat2));
lat2 = rad2deg(lat2);
lon2 = rad2deg(lon2);
newLocation.setLatitude(lat2);
newLocation.setLongitude(lon2);
return newLocation;
}
private: static double rad2deg(double radians)
{
static double ratio = (double)(180.0/3.141592653589793238);
return radians * ratio;
}
private: static double deg2rad(double radians)
{
static double ratio = (double)(180.0/3.141592653589793238);
return radians/ratio;
}
/*
public: static void testCoarse()
{
Location vancouver(49.2445, -123.099146);
Location vancouver2 = makeLocationCoarse(vancouver);
Location korea(37.423938, 126.692488);
Location korea2 = makeLocationCoarse(korea);
Location hiroshima(34.3937, 132.464);
Location hiroshima2 = makeLocationCoarse(hiroshima);
Location zagreb(45.791958, 15.935786);
Location zagreb2 = makeLocationCoarse(zagreb);
Location anchorage(61.367778, -149.900208);
Location anchorage2 = makeLocationCoarse(anchorage);
}*/
};
如果假設地球是一個球體（基本上適用於這個問題），這非常類似於上一個問題 Rounding Lat and Long to Show Approximate Location in Google Maps
，那麼你只需要計算一個位置，這是從一個特定的角度距離給予經緯度。選擇距離和（隨機）方向，並使用距離公式計算新位置。
這裏有相反的問題的很好的討論（二緯度/經度點之間的距離）：http://mathforum.org/library/drmath/view/51756.html
它應該是相對直接從那裏去尋找一個點具有指定距離指定點。
我嘗試implemente在Ruby，但在我的情況的解決方案，粗協調VS現實有巨大的差異。粗略座標僅在經緯度變化時變化，但當緯度保持不變並且長時間移動時，粗略保持不變。如果有人可以檢查下面的代碼，也許我做了一個錯誤的編碼。
class CoarseLocation
AREA_LIMIT = 1000
class << self
def make_location_coarse(lat, lon)
if lat.nil? && lon.nil?
raise InvalidParamsError
end
location = [lat.to_f, lat.to_f]
new_location_up = get_location_by_offset(location, AREA_LIMIT, 0)
granularityLat = location[0] - new_location_up[0]
if granularityLat < 0
granularityLat = -granularityLat
end
new_location_right = get_location_by_offset(location, AREA_LIMIT, 1.57079633)
granularityLon = location[1] - new_location_right[1]
if(granularityLon < 0)
granularityLon = -granularityLon
end
course_lat = location[0]
course_lon = location[1]
if(granularityLat == 0.0) || (granularityLon == 0.0)
course_lat = 0
course_lon = 0
else
course_lat = (course_lat/granularityLat).to_i * granularityLat
course_lon = (course_lon/granularityLon).to_i * granularityLon
end
[course_lat, course_lon]
end
def get_location_by_offset(location, offset, angle)
lat_radius = location[0] * Math::PI/180
lon_radius = location[1] * Math::PI/180
distance = (offset/1000).to_f
earth_radius = 6371
lat_radius_1 = (Math::asin(Math::sin(lat_radius) * Math::cos(distance/earth_radius) + Math::cos(lat_radius) * Math::sin(distance/earth_radius) * Math::cos(angle))).to_f
lon_radius_1 = (lon_radius + Math::atan2(Math::sin(angle)*Math::sin(distance/earth_radius)*Math::cos(lat_radius), Math::cos(distance/earth_radius) - Math::sin(lat_radius)*Math::sin(lat_radius_1))).to_f
new_lat = lat_radius_1 * 180/Math::PI
new_lon = lon_radius_1 * 180/Math::PI
return [new_lat.to_f, new_lon.to_f]
end
end
end
位置字段總是包含2個元素的陣列，其中[0]是lat，[1]是long。
請記住，相似的長度有所不同。我的意思是在赤道附近（緯度〜0°），一個單一的經度是一個更大的距離，然後是靠近兩極的一個單一的經度。在極點（緯度+/- 90°）處，經度甚至失去了它的意義。 – Plap