给定一组纬度和经度点,我如何计算该集合中心点的纬度和经度(也就是将视图集中在所有点上的一个点)?
编辑:我使用过的Python解决方案:
Convert lat/lon (must be in radians) to Cartesian coordinates for each location.
X = cos(lat) * cos(lon)
Y = cos(lat) * sin(lon)
Z = sin(lat)
Compute average x, y and z coordinates.
x = (x1 + x2 + ... + xn) / n
y = (y1 + y2 + ... + yn) / n
z = (z1 + z2 + ... + zn) / n
Convert average x, y, z coordinate to latitude and longitude.
Lon = atan2(y, x)
Hyp = sqrt(x * x + y * y)
Lat = atan2(z, hyp)
发布于 2013-01-09 16:46:04
谢谢!下面是使用degrees的OP解决方案的C#版本。它利用System.Device.Location.GeoCoordinate类
public static GeoCoordinate GetCentralGeoCoordinate(
IList<GeoCoordinate> geoCoordinates)
{
if (geoCoordinates.Count == 1)
{
return geoCoordinates.Single();
}
double x = 0;
double y = 0;
double z = 0;
foreach (var geoCoordinate in geoCoordinates)
{
var latitude = geoCoordinate.Latitude * Math.PI / 180;
var longitude = geoCoordinate.Longitude * Math.PI / 180;
x += Math.Cos(latitude) * Math.Cos(longitude);
y += Math.Cos(latitude) * Math.Sin(longitude);
z += Math.Sin(latitude);
}
var total = geoCoordinates.Count;
x = x / total;
y = y / total;
z = z / total;
var centralLongitude = Math.Atan2(y, x);
var centralSquareRoot = Math.Sqrt(x * x + y * y);
var centralLatitude = Math.Atan2(z, centralSquareRoot);
return new GeoCoordinate(centralLatitude * 180 / Math.PI, centralLongitude * 180 / Math.PI);
}
发布于 2013-09-05 04:59:29
我发现这篇文章非常有用,所以这里是用PHP编写的解决方案。我已经成功地使用了这个,只是想节省另一个开发人员的时间。
/**
* Get a center latitude,longitude from an array of like geopoints
*
* @param array data 2 dimensional array of latitudes and longitudes
* For Example:
* $data = array
* (
* 0 = > array(45.849382, 76.322333),
* 1 = > array(45.843543, 75.324143),
* 2 = > array(45.765744, 76.543223),
* 3 = > array(45.784234, 74.542335)
* );
*/
function GetCenterFromDegrees($data)
{
if (!is_array($data)) return FALSE;
$num_coords = count($data);
$X = 0.0;
$Y = 0.0;
$Z = 0.0;
foreach ($data as $coord)
{
$lat = $coord[0] * pi() / 180;
$lon = $coord[1] * pi() / 180;
$a = cos($lat) * cos($lon);
$b = cos($lat) * sin($lon);
$c = sin($lat);
$X += $a;
$Y += $b;
$Z += $c;
}
$X /= $num_coords;
$Y /= $num_coords;
$Z /= $num_coords;
$lon = atan2($Y, $X);
$hyp = sqrt($X * $X + $Y * $Y);
$lat = atan2($Z, $hyp);
return array($lat * 180 / pi(), $lon * 180 / pi());
}
发布于 2015-05-04 23:10:05
非常有用的帖子!我已经在JavaScript中实现了这一点,这里是我的代码。我已经成功地使用了这一点。
function rad2degr(rad) { return rad * 180 / Math.PI; }
function degr2rad(degr) { return degr * Math.PI / 180; }
/**
* @param latLngInDeg array of arrays with latitude and longtitude
* pairs in degrees. e.g. [[latitude1, longtitude1], [latitude2
* [longtitude2] ...]
*
* @return array with the center latitude longtitude pairs in
* degrees.
*/
function getLatLngCenter(latLngInDegr) {
var LATIDX = 0;
var LNGIDX = 1;
var sumX = 0;
var sumY = 0;
var sumZ = 0;
for (var i=0; i<latLngInDegr.length; i++) {
var lat = degr2rad(latLngInDegr[i][LATIDX]);
var lng = degr2rad(latLngInDegr[i][LNGIDX]);
// sum of cartesian coordinates
sumX += Math.cos(lat) * Math.cos(lng);
sumY += Math.cos(lat) * Math.sin(lng);
sumZ += Math.sin(lat);
}
var avgX = sumX / latLngInDegr.length;
var avgY = sumY / latLngInDegr.length;
var avgZ = sumZ / latLngInDegr.length;
// convert average x, y, z coordinate to latitude and longtitude
var lng = Math.atan2(avgY, avgX);
var hyp = Math.sqrt(avgX * avgX + avgY * avgY);
var lat = Math.atan2(avgZ, hyp);
return ([rad2degr(lat), rad2degr(lng)]);
}
https://stackoverflow.com/questions/6671183
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