如何根据Android中当前位置的距离对地理点进行排序

问题描述 投票:3回答:4

我有一个“Place”对象,每个对象都有一个LatLng坐标:

import com.google.android.gms.maps.model.LatLng;

public class Place{
    public String name;
    public LatLng latlng;

    public Restaurant(String name, LatLng latlng) {
        this.name = name;
        this.latlng = latlng;
    }
}

我有这些地方的ArrayList,如下所示:

    ArrayList<Place> places = new ArrayList<Place>();
    places.add("Place 1", LatLng(90.0,90.0));
    places.add("Place 2", LatLng(93.0,93.0));
    places.add("Place 3", LatLng(83.0,92.0));
    places.add("Place 4", LatLng(93.0,91.0));

我有“我的”LatLng:

    LatLng myLocation = new LatLng(10.0,10.0);

如何根据离我最近的方式对这些物体进行排序?谢谢您的帮助

android geolocation latitude-longitude
4个回答
6
投票

从使用this answer的@shieldstroy发布的问题中获取Great Circle Distance的算法,我得到了这个例子。

这是Comparator

public class SortPlaces implements Comparator<Place> {
    LatLng currentLoc;

    public SortPlaces(LatLng current){
        currentLoc = current;
    }
    @Override
    public int compare(final Place place1, final Place place2) {
        double lat1 = place1.latlng.latitude;
        double lon1 = place1.latlng.longitude;
        double lat2 = place2.latlng.latitude;
        double lon2 = place2.latlng.longitude;

        double distanceToPlace1 = distance(currentLoc.latitude, currentLoc.longitude, lat1, lon1);
        double distanceToPlace2 = distance(currentLoc.latitude, currentLoc.longitude, lat2, lon2);
        return (int) (distanceToPlace1 - distanceToPlace2);
    }

    public double distance(double fromLat, double fromLon, double toLat, double toLon) {
        double radius = 6378137;   // approximate Earth radius, *in meters*
        double deltaLat = toLat - fromLat;
        double deltaLon = toLon - fromLon;
        double angle = 2 * Math.asin( Math.sqrt(
                Math.pow(Math.sin(deltaLat/2), 2) +
                        Math.cos(fromLat) * Math.cos(toLat) *
                                Math.pow(Math.sin(deltaLon/2), 2) ) );
        return radius * angle;
    }
}

这是高级代码,我把它放在onCreate()中:

        //My location, San Francisco
        double lat = 37.77657;
        double lng = -122.417506;
        LatLng latLng = new LatLng(lat, lng);

        //set up list
        ArrayList<Place> places = new ArrayList<Place>();

        places.add(new Place("New York", new LatLng(40.571256,73.98369)));
        places.add(new Place("Colorado", new LatLng(39.260658,-105.101615)));
        places.add(new Place("Los Angeles", new LatLng(33.986816,118.473819)));

        for (Place p: places){
            Log.i("Places before sorting", "Place: " + p.name);
        }

        //sort the list, give the Comparator the current location
        Collections.sort(places, new SortPlaces(latLng));

        for (Place p: places){
            Log.i("Places after sorting", "Place: " + p.name);
        }

这是日志输出:

04-17 23:04:16.074  12963-12963/com.maptest.daniel.maptest I/Places before sorting﹕ Place: New York
04-17 23:04:16.074  12963-12963/com.maptest.daniel.maptest I/Places before sorting﹕ Place: Colorado
04-17 23:04:16.074  12963-12963/com.maptest.daniel.maptest I/Places before sorting﹕ Place: Los Angeles
04-17 23:04:16.074  12963-12963/com.maptest.daniel.maptest I/Places after sorting﹕ Place: Los Angeles
04-17 23:04:16.074  12963-12963/com.maptest.daniel.maptest I/Places after sorting﹕ Place: Colorado
04-17 23:04:16.074  12963-12963/com.maptest.daniel.maptest I/Places after sorting﹕ Place: New York

1
投票

为了计算距离,可以使用不同的方法。一个非常简单的是Haversine Formula(http://rosettacode.org/wiki/Haversine_formula#Java)。一个更准确的计算是Vincenty公式。如果这两个位置不远,那么Haversine解决方案就足够了。

计算距离后,您只需使用比较器对数组进行排序,例如:

Collections.sort(places, new Comparator<Place>() {
    public int compare(Place p1, Place p2) {
        return Double.compare(p1.getDistance(), p2.getDistance());
    }
});

1
投票

当您获得当前位置时,您可以通过计算行驶距离(最适合餐馆等场所)进行排序,如下所示:

  1. 计算每个对象的距离 http://maps.googleapis.com/maps/api/directions/json?origin="+yourLat+","+yourLong+"&destination="+toLat+","+toLong+"&sensor=false&mode=DRIVING
  2. 计算每个距离后,对这些距离应用一些简单的排序算法。

0
投票

你可以使用

distanceInMeters = (loc1.distanceTo(loc2));

从谷歌地图API,然后将结果添加到TreeMap中的键

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