我知道，我所要求的内容存在于 Google Map 中，但我正在使用 Xamarin.Forms.Map，所以..我必须自己制作。
但是，我知道如何获得我的点的中心，即POI(兴趣点)，但我不知道如何确定相机的变焦..
我在网上搜索了from this post ，我被重定向到 Haversine的算法。
但是，我尝试了给出的代码，但它不起作用..我知道如何找到 POI，即 2 个最远的点，但我无法确定缩放..
请问有什么想法吗？ :/
注意:如果您想了解我尝试过的内容，可以使用代码
#region Camera focus method
private static void OnCustomPinsPropertyChanged(BindableObject bindable, object oldValue, object newValue)
{
CustomMap customMap = ((CustomMap)bindable);
if (customMap.CameraFocusParameter == CameraFocusReference.OnPins)
{
List<Position> PositionPins = new List<Position>();
bool onlyOnePointPresent;
foreach (CustomPin pin in (newValue as List<CustomPin>))
{
PositionPins.Add(pin.Position);
}
Position CentralPosition = GetCentralPosition(PositionPins);
if (PositionPins.Count > 1)
{
Position[] FarestPoints = GetTwoFarestPointsOfCenterPointReference(PositionPins, CentralPosition);
customMap.CameraFocus = GetPositionAndZoomLevelForCameraAboutPositions(FarestPoints);
onlyOnePointPresent = false;
}
else
{
customMap.CameraFocus = new CameraFocusData() { Position = CentralPosition };
onlyOnePointPresent = true;
}
customMap.MoveToRegion(MapSpan.FromCenterAndRadius(customMap.CameraFocus.Position,
(!onlyOnePointPresent) ? (customMap.CameraFocus.Distance) : (new Distance(5))));
}
}
public static Position GetCentralPosition(List<Position> positions)
{
if (positions.Count == 1)
{
foreach (Position pos in positions)
{
return (pos);
}
}
double lat = 0;
double lng = 0;
foreach (var pos in positions)
{
lat += pos.Latitude;
lng += pos.Longitude;
}
var total = positions.Count;
lat = lat / total;
lng = lng / total;
return new Position(lat, lng);
}
public class DataCalc
{
public Position Pos { get; set; }
public double Distance { get; set; }
}
public static Position[] GetTwoFarestPointsOfCenterPointReference(List<Position> farestPosition, Position centerPosition)
{
Position[] FarestPos = new Position[2];
List<DataCalc> dataCalc = new List<DataCalc>();
Debug.WriteLine("So the center is on [{0}]/[{1}]", centerPosition.Latitude, centerPosition.Longitude);
foreach (Position pos in farestPosition)
{
dataCalc.Add(new DataCalc()
{
Pos = pos,
Distance = Math.Sqrt(Math.Pow(pos.Latitude - centerPosition.Latitude, 2) + Math.Pow(pos.Longitude - centerPosition.Longitude, 2))
});
}
DataCalc First = new DataCalc() { Distance = 0 };
foreach (DataCalc dc in dataCalc)
{
if (dc.Distance > First.Distance)
{
First = dc;
}
}
Debug.WriteLine("The farest one is on [{0}]/[{1}]", First.Pos.Latitude, First.Pos.Longitude);
DataCalc Second = new DataCalc() { Distance = 0 };
foreach (DataCalc dc in dataCalc)
{
if (dc.Distance > Second.Distance
&& (dc.Pos.Latitude != First.Pos.Latitude && dc.Pos.Longitude != First.Pos.Longitude))
{
Second = dc;
}
}
Debug.WriteLine("the second is on [{0}]/[{1}]", Second.Pos.Latitude, Second.Pos.Longitude);
FarestPos[0] = First.Pos;
FarestPos[1] = Second.Pos;
return (FarestPos);
}
public class CameraFocusData
{
public Position Position { get; set; }
public Distance Distance { get; set; }
}
//HAVERSINE
public static CameraFocusData GetPositionAndZoomLevelForCameraAboutPositions(Position[] FarestPoints)
{
double earthRadius = 6371000; //metros
Position pos1 = FarestPoints[0];
Position pos2 = FarestPoints[1];
double latitud1Radianes = pos1.Latitude * (Math.PI / 180.0);
double latitud2Radianes = pos2.Latitude * (Math.PI / 180.0);
double longitud1Radianes = pos2.Longitude * (Math.PI / 180.0);
double longitud2Radianes = pos2.Longitude * (Math.PI / 180.0);
double deltaLatitud = (pos2.Latitude - pos1.Latitude) * (Math.PI / 180.0);
double deltaLongitud = (pos2.Longitude - pos1.Longitude) * (Math.PI / 180.0);
double sum1 = Math.Sin(deltaLatitud / 2) * Math.Sin(deltaLatitud / 2);
double sum2 = Math.Cos(latitud1Radianes) * Math.Cos(latitud2Radianes) * Math.Sin(deltaLongitud / 2) * Math.Sin(deltaLongitud / 2);
var a = sum1 + sum2;
var c = 2 * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1 - a));
var distance = earthRadius * c;
/* lt is deltaLatitud
* lng is deltaLongitud*/
var Bx = Math.Cos(latitud2Radianes) * Math.Cos(deltaLongitud);
var By = Math.Cos(latitud2Radianes) * Math.Sin(deltaLongitud);
var lt = Math.Atan2(Math.Sin(latitud1Radianes) + Math.Sin(latitud2Radianes),
Math.Sqrt((Math.Cos(latitud1Radianes) + Bx) * (Math.Cos(latitud2Radianes) + Bx) + By * By));//Latitud del punto medio
var lng = longitud1Radianes + Math.Atan2(By, Math.Cos(longitud1Radianes) + Bx);//Longitud del punto medio
Debug.WriteLine("the final pos of the camera is on [{0}]/[{1}]", lt, lng);
return (new CameraFocusData() { Position = new Position(lt, lng), Distance = new Distance(distance + 0.2) });
}
#endregion
最佳答案
然后我找到了解决方案，有它的代码，它已被写入以放入您的自定义 map 中。
在这里，private static void OnCustomPinsPropertyChanged(BindableObject bindable, object oldValue, object newValue)是我的 List<CustomPins> 调用的方法但您可以使用不同的方法。
public static readonly BindableProperty CustomPinsProperty =
BindableProperty.Create(nameof(CustomPins), typeof(IList<CustomPin>), typeof(CustomMap), null,
propertyChanged: OnCustomPinsPropertyChanged);
此外，您可以添加用户的纬度/经度，我没有这样做，因为我的需要没有用户的位置:)。
最后，你可以为缩放添加一个乘数，我的意思是，你可以说，嗯，缩放对我来说太远了，那么好吧，像我一样，将 double distance 相乘。值如 0.7或0.6 :)
#region Camera focus definition
public class CameraFocusData
{
public Position Position { get; set; }
public Distance Distance { get; set; }
}
private static void OnCustomPinsPropertyChanged(BindableObject bindable, object oldValue, object newValue)
{
CustomMap customMap = ((CustomMap)bindable);
if (customMap.CameraFocusParameter == CameraFocusReference.OnPins)
{
List<double> latitudes = new List<double>();
List<double> longitudes = new List<double>();
foreach (CustomPin pin in (newValue as List<CustomPin>))
{
latitudes.Add(pin.Position.Latitude);
longitudes.Add(pin.Position.Longitude);
}
double lowestLat = latitudes.Min();
double highestLat = latitudes.Max();
double lowestLong = longitudes.Min();
double highestLong = longitudes.Max();
double finalLat = (lowestLat + highestLat) / 2;
double finalLong = (lowestLong + highestLong) / 2;
double distance = DistanceCalculation.GeoCodeCalc.CalcDistance(lowestLat, lowestLong, highestLat, highestLong, DistanceCalculation.GeoCodeCalcMeasurement.Kilometers);
customMap.MoveToRegion(MapSpan.FromCenterAndRadius(new Position(finalLat, finalLong), Distance.FromKilometers(distance * 0.7)));
}
}
private class DistanceCalculation
{
public static class GeoCodeCalc
{
public const double EarthRadiusInMiles = 3956.0;
public const double EarthRadiusInKilometers = 6367.0;
public static double ToRadian(double val) { return val * (Math.PI / 180); }
public static double DiffRadian(double val1, double val2) { return ToRadian(val2) - ToRadian(val1); }
public static double CalcDistance(double lat1, double lng1, double lat2, double lng2)
{
return CalcDistance(lat1, lng1, lat2, lng2, GeoCodeCalcMeasurement.Miles);
}
public static double CalcDistance(double lat1, double lng1, double lat2, double lng2, GeoCodeCalcMeasurement m)
{
double radius = GeoCodeCalc.EarthRadiusInMiles;
if (m == GeoCodeCalcMeasurement.Kilometers) { radius = GeoCodeCalc.EarthRadiusInKilometers; }
return radius * 2 * Math.Asin(Math.Min(1, Math.Sqrt((Math.Pow(Math.Sin((DiffRadian(lat1, lat2)) / 2.0), 2.0) + Math.Cos(ToRadian(lat1)) * Math.Cos(ToRadian(lat2)) * Math.Pow(Math.Sin((DiffRadian(lng1, lng2)) / 2.0), 2.0)))));
}
}
public enum GeoCodeCalcMeasurement : int
{
Miles = 0,
Kilometers = 1
}
}
#endregion
玩得开心!
2018 年 12 月更新
我开发了一个很久以前在我的存储库上托管的解决方案，与其他 POC https://github.com/Emixam23/XamarinByEmixam23/tree/master/Detailed%20Part/Controls/Map/MapPinsProject
关于c# - 确定缩放级别以覆盖有关纬度/经度的所有标记，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/39536580/