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记一次 Sedona(GeoSpark) 空间计算优化

开发技术 开发技术 2022-09-07 次浏览

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
  )
程序员灯塔
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