Mongo 地理位置查询：海量密集点转换成聚合信息

通俗来说：将地图上的海量密集点通过网格分割的方式实现聚合；

需求：用mongo实现设备地理位置聚合查询：多边形，矩形查询;

背景：上万设备数据量

目的：分享Mongo地理位置查询，以及文末对在此之前的两种实现方式做分析比较，纠正一些开发中的错误认知；

1、自定义：数据库查询

//多边形   $polygon 
@Query(value = "{'position': {$exists: true}, $and: [{'position': {$geoWithin: { $polygon : ?0 }}}, {'deletedAt': null}]}" , fields = "{ 'position': 1 }")List<Thing> findByProvinceBoundary(double[][] provinceBoundary);//矩形 $box@CountQuery("{$and: [{'position': {$exists: true}}, {'deletedAt': null},"+ "{'position': {$geoWithin: { $box: [?0, ?1] }}}]}")long countByProvinceBoundaryOrBoundingBox(double[] bottomLeft, double[] topRight);

注意：MongoDB 不支持在一个查询中同时使用 $polygon 和 $box；

2、如果需要求中心点，可以使用聚合查询，实现加权平均权重点

 @Aggregation(pipeline = {"{ $match: { $and: [ { 'position': { $exists: true } }, { 'deletedAt': null }, { 'position': { $geoWithin: { $box: [?0, ?1] } } } ] } }","{ $group: { _id: null, longitude: { $avg: '$position.longitude' }, latitude: { $avg: '$position.latitude' } } }"})GeoPoint findWeightedCenter(double[] bottomLeft, double[] topRight);

3、完整设备位置聚合信息查询：最佳方案

 @Overridepublic ThingGeo getAggregatedThingGeo(ThingGeoReqDTO reqDTO) {Area area = areaRepository.getAreaByCode();//行政编码//1.行政编码区域的中心点，查询没有位置的设备总数：JSONObject properties = area.getBound().getJSONArray("features").getJSONObject(0).getJSONObject("properties");JSONArray centerPosition= properties.getJSONArray("center"); //中心点位置double centerLon = centerPosition.getDouble(0);double centerLat = centerPosition.getDouble(1);GeoPoint centerPoint = new GeoPoint(centerLon, centerLat);long noGeoThingCount = thingRepository.countByNoGeoPosition();GridCellThing noGeoThings = new GridCellThing(centerPoint,noGeoThingCount);//2.网格查询有位置信息的设备总数以及权重点double[] topRight = reqDTO.getTopRight();double[] bottomLeft = reqDTO.getBottomLeft();// 计算X和Y的差值(视图长和宽)double deltaX = topRight[0] - bottomLeft[0];double deltaY = topRight[1] - bottomLeft[1];// 计算X和Y的平均值double avgX = deltaX / 4;double avgY = deltaY / 4;// 使用右上角作为起始点double x = topRight[0];double y = topRight[1];List<GridCellThing> gridCellThings = new ArrayList<>();// 循环生成4*4=16网格for (int a = 0; a < 4; a++) {for (int i = 0; i < 4; i++) {// 计算网格边界double minX = x - (i + 1) * avgX;double maxX = x - i * avgX;double minY = y - (a + 1) * avgY;double maxY = y - a * avgY;//小网格：两个对角经纬度double[] boxTopRight = new double[] {maxX, maxY};double[] boxBottomLeft = new double[] {minX, minY};long boxCount = thingRepository.countByBoundingBox(boxBottomLeft,boxTopRight);if (boxCount > 0) {GeoPoint center = thingRepository.findWeightedCenter(boxBottomLeft, boxTopRight);GeoPoint boxCenter = new GeoPoint(center.getLongitude(),center.getLatitude());GridCellThing gridCellThing = new GridCellThing();gridCellThing.setThingCount(boxCount);gridCellThing.setPosition(boxCenter);gridCellThings.add(gridCellThing);}}}ThingGeo thingGeo = new ThingGeo();thingGeo.setGridCellList(gridCellThings);thingGeo.setNoGeoThings(noGeoThings);return thingGeo;}

4、在此之前的踩坑错误实现代码：在数据量多的时候，导致内存溢出；

因为拿到几万条设备信息导致内存溢出；

 public static List<GridCellThing> getGridCellThings(ThingGeoReqDTO reqVO, List<Thing> things) {double[] topRight = reqVO.getTopRight();double[] bottomLeft = reqVO.getBottomLeft();// 计算X和Y的差值(视图长和宽)double deltaX = topRight[0] - bottomLeft[0];double deltaY = topRight[1] - bottomLeft[1];// 计算X和Y的平均值double avgX = deltaX / 4;double avgY = deltaY / 4;// 使用右上角作为起始点double x = topRight[0];double y = topRight[1];List<GridCellThing> gridCellThings = new ArrayList<>();// 循环生成4*4=16网格for (int a = 0; a < 4; a++) {for (int i = 0; i < 4; i++) {GridCellThing gridCellThing = new GridCellThing();// 计算网格边界double minX = x - (i + 1) * avgX;double maxX = x - i * avgX;double minY = y - (a + 1) * avgY;double maxY = y - a * avgY;double centerTotalX = 0;double centerTotalY = 0;int count = 0;// 遍历设备列表for (Thing thing : things) {double longitude = thing.getPosition().getLongitude();double latitude = thing.getPosition().getLatitude();// 检查设备是否在当前网格内，根据最大值最小值区分来做到去重边界相交的设备;if (longitude > minX && longitude <= maxX && latitude > minY && latitude <= maxY) {count++;//得到有效设备的经纬度总和centerTotalX += longitude;centerTotalY += latitude;}}// 如果有符合条件的设备，则记录到响应对象列表中if (count > 0) {//分别把经度总和、纬度总和 除以count(网格区域内设备总数)得到经纬度的加权平均值double centerLongitude = centerTotalX / count;double centerLatitude = centerTotalY / count;GeoPoint geoPoint = new GeoPoint();geoPoint.setLongitude(centerLongitude);geoPoint.setLatitude(centerLatitude);gridCellThing.setPosition(geoPoint);gridCellThing.setThingCount(count);gridCellThings.add(gridCellThing);}}}return gridCellThings;}