项目需求需要空间计算能力，开始选型Sedona（GeoSpark）来完成，
需求需要每一条数据在满足某条件的情况下，去查找某张表进行空间匹配，找到离这个点（point）最近的一条道路（lineString）
第一个方案: 使用sedona来使用临近道路的判断
由于sedona本质还是使用spark的能力，所以遵循spark的开发规则，不能在rdd.map 里面干活，sedona也不支持批量查，只能一条一条匹配。 伪代码如下
val spatial_sql = """ | select | ST_GeomFromWKT(geom) geom, name, adcode | from ods.ods_third_party_road_data |""".stripMargin val third_party_road_df = spark.sql(spatial_sql).toDF() aoi_day_s_df.rdd.collect().par.map(row => { val tmp_location = row.getAs[String]("poi_location") val near_street = spatialQueryStreet(third_party_road_df, city_code, tmp_location) println(near_street) ... ) def spatialQueryStreet(third_party_road_df:DataFrame, city_code:String, location: String): String = { val frame = third_party_road_df.where("adcode = '%s'".format(city_code)).toDF() val tp_road_spatial_rdd = Adapter.toSpatialRdd(frame, "geom") tp_road_spatial_rdd.buildIndex(IndexType.RTREE, false) val geometryFactory = new GeometryFactory() val x = location.substring(location.indexOf("(") + 1, location.indexOf(" ")) val y = location.substring(location.indexOf(" ") + 1, location.indexOf(")")) val pointObject = geometryFactory.createPoint(new Coordinate(x.toDouble, y.toDouble)) val usingIndex = true val result = KNNQuery.SpatialKnnQuery(tp_road_spatial_rdd, pointObject, 1, usingIndex) if (result.isEmpty) { return "" } else { val dst = result.get(0) //System.out.println("==== dst.getUserData: " + dst.getUserData.toString) val strings = dst.getUserData.toString.split("\t") val near_street = strings(0) //System.out.println("==== near_street: " + near_street) near_street }
结果效率不高，因为每条数据都要匹配，sedona又不能在rdd.map中使用，所以必须先collect().map,这就不能利用到spark多节点并行的特性； 2. 每条数据都基于third_party_road_df创建了空间索引来查，效率更低了（如果只有一条数据还勉强可以接受）
方案2： 改sedona为JTS来处理，jts直接创建rtree，可以在rdd.map中处理，而且创建速度也更快一些，效率更高了
伪代码如下
poi_build_aoi_aoi_day_s_df.rdd.map(row => { val tmp_location = row.getAs[String]("poi_location") val rtree = createRtree(model_list) near_street = spatialQueryStreet(rtree, tmp_location) println(near_street) ... ) def createRtree(third_party_road_list: Array[ThirdPartyModel]): STRtree = { val rtree = new STRtree() for (model <- third_party_road_list) { val geom = model.geometry geom.setUserData(model.name) rtree.insert(geom.getEnvelopeInternal, model.geometry) } rtree.build() rtree } def spatialQueryStreet(rtree: STRtree, location: String): String = { if (rtree == null) { "" } val geometryFactory = new GeometryFactory() val x = location.substring(location.indexOf("(") + 1, location.indexOf(" ")) val y = location.substring(location.indexOf(" ") + 1, location.indexOf(")")) val pointObject = geometryFactory.createPoint(new Coordinate(x.toDouble, y.toDouble)) val result = rtree.nearestNeighbour(pointObject.getEnvelopeInternal, pointObject, new GeometryItemDistance()) val name = result.asInstanceOf[Geometry].getUserData.asInstanceOf[String] println(s"nearestNeighbour name: $name") name }
通过这次修改，由原来跑3个小时（甚至更多）的任务在15分钟内就跑完了
PS： 经尝试rtree 不能通过广播变量发送出去，会报序列化异常
其实还可以再优化一下，上面每条数据还是创建了一次rtree, 可以改为mapPartition，然后只建一次rtree, 数据量大时效果更佳
aoi_day_s_df.rdd.mapPartitions(iterator => { // rtree 放到iterator.map 外面创建，搞一次就ok了，更快（不过我没有试验，应该是百分百可行的） val rtree = createRtree(model_list) val seq = iterator.map(row => { val tmp_location = row.getAs[String]("poi_location") near_street = spatialQueryStreet(rtree, tmp_location) println(near_street) ... ) seq )
