找到最近点（几何点）

Question

当我有几何图形时，我想计算离起点（参考）最近的点 point。

出于这个原因，我正在使用 ports.shp 文件。

该代码在 time.But 的大部分时间都有效，有时 returns 无效 minDistPoint is null。我不确定初始化什么值 minDist.

public Point findNearestPoint( Point p, SimpleFeatureCollection features ) throws FactoryException, TransformException {
   Point destination = null;
   double minDist = 10.0e+6;
   double distance = 0;
   Point minDistPoint = null;
   try( SimpleFeatureIterator itr = features.features()) {
      while( itr.hasNext()) {
         SimpleFeature feature = itr.next();
         final String EPSG4326 =
            "GEOGCS[\"WGS 84\",DATUM[\"WGS_1984\","+
            "SPHEROID[\"WGS 84\",6378137,298.257223563,"+
            "AUTHORITY[\"EPSG\",\"7030\"]],"+
            "AUTHORITY[\"EPSG\",\"6326\"]]," + 
            "PRIMEM[\"Greenwich\",0,AUTHORITY[\"EPSG\",\"8901\"]],"+
            "UNIT[\"degree\", " +"0.01745329251994328,"+
            "AUTHORITY[\"EPSG\",\"9122\"]],"+
            "AUTHORITY[\"EPSG\",\"4326\"]]";
         CoordinateReferenceSystem crs = CRS.parseWKT(EPSG4326);    
         destination = (Point) feature.getDefaultGeometry(); 
         GeodeticCalculator gc = new GeodeticCalculator(crs); 
         gc.setStartingPosition(
            JTS.toDirectPosition( p.getCoordinate(), crs));
         gc.setDestinationPosition(
            JTS.toDirectPosition( dest.getCoordinate(), crs));
         distance = gc.getOrthodromicDistance();
         if( distance < minDist ) {
            minDist = distance;
            minDistPoint = destination;
            lastMatched = feature;
         }
      }
   }
   int totalmeters = (int) minDist;
   int km = totalmeters / 1000;
   int meters = totalmeters - (km * 1000);
   float remaining_cm = (float) (minDist - totalmeters) * 10000;
   remaining_cm = Math.round(remaining_cm);
   float cm = remaining_cm / 100;
   System.out.println(
      "Distance = " + km + "km " + meters + "m " + cm + "cm");
   if( minDistPoint == null ) {
      return null;
   }
   return minDistPoint;
}

Answer 1

对于每一个寻找最小值的算法，您都应该始终将初始值设置为集合中的第一个值或可实现的最大值。

在您的情况下，由于您使用的是双精度值，因此可以使用 Double.POSITIVE_INFINITY 作为初始值，这将使您至少可以确保 if 条件第一次为真.

Answer 2

评论中给出了计算机科学答案，Double.PositiveInfinity比任何数字都大。地理空间答案是在一个球体上没有两点相距超过 pi*r。对于地球来说，一个好的上限是 3.14*6400 公里。使用正无穷大。

找到最近点（几何点）

find nearest point (geometry point)

java

gis

geotools