我是Apache Spark Streaming的新手。我正在开发一个spark流媒体应用程序,以找到最短的路径,并再次将该路径发送回客户端。我已经写好了数据处理的代码,但我有一个问题,我怎么能把我的结果再次送回客户端这里是我的代码。
import networkx as nx
from pyspark import SparkConf,SparkContext
from pyspark.streaming import StreamingContext
TCP_IP = "127.0.0.1"
TCP_PORT = 5000
# Creating a Spark Configuration
conf=SparkConf()
conf.setAppName('ShortestPathApp')
sc= SparkContext(conf)
ssc= StreamingContext(sc,2)
def shortestPath(line):
# get the values from rdd
vehicleId = line[0]
source = line[1]
destination = line[2]
deadline = line[3]
# find shortest path
shortest = nx.dijkstra_path(G, source, destination)
# receive from Socket
dataStream =ssc.socketTextStream(TCP_IP,TCP_PORT)
vehicle_data = dataStream.map(lambda line: line.split(" "))
vehicle_data.foreachRDD(lambda rdd: rdd.foreach(shortestPath))
ssc.start()
ssc.awaitTermination()
我怎么能把数据发回客户端呢?
使用 StreamingContext
将输出数据以流的形式推送回目的地,你可以创建如下方法。
# Lazily instantiated global instance of SparkSession
def getSparkSessionInstance(sparkConf):
if ("sparkSessionSingletonInstance" not in globals()):
globals()["sparkSessionSingletonInstance"] = SparkSession \
.builder \
.config(conf=sparkConf) \
.getOrCreate()
return globals()["sparkSessionSingletonInstance"]
sparkSess = getSparkSessionInstance(rdd.context.getConf())
vehicle_data_df = sparkSess.createDataFrame(vehicle_data)
vehicle_data_df.writeStream\
.format("socket")\
.option("host",TCP_OUTPUT_IP) //Output socket IP address
.option("port",TCP_OUTPUT_PORT) //Output socket port
.outputMode('append')\
.start()\
.awaitTermination()