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py4j.Py4JException:方法socketTextStream不存在

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我是Spark Streaming的新手 . 在PyCharm中使用PySpark我无法通过socketTextStream初始化 .

def start_streaming (self):
    sp = SparkContext('local[2]', 'streamingTest')
    stream = StreamingContext(sp, 1)
    **items = stream.socketTextStream('localhost', '9009')**
    print(items)
    lines = items.flatMap(lambda line : line.split('\n'))

我收到以下错误 . 关于问题是什么的任何想法?

Traceback(最近一次调用最后一次):文件“/Users/egorkorneev/PycharmProjects/BasicStreaming/SparkConsumer.py”,第75行,在dc.start_streaming()文件“/Users/egorkorneev/PycharmProjects/BasicStreaming/SparkConsumer.py”中,第40行,在start_streaming items = stream.socketTextStream('localhost','9009')文件“/Users/egorkorneev/rnd/spark-1.6.1-bin-hadoop2.6/python/pyspark/streaming/context.py” ,第352行,在socketTextStream中返回DStream(self._jssc.socketTextStream(hostname,port,jlevel),self,File“/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/py4j /java_gateway.py“,第933行,在呼叫回答中,self.gateway_client,self.target_id,self.name)文件”/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/ py4j / protocol.py“,第316行,采用get_return_value格式(target_id,” . “,名称,值))py4j.protocol.Py4JError:调用o19.socketTextStream时发生错误 . 跟踪:py4j.Py4JException:方法socketTextStream([类java.lang.String,类java.lang.String,类org.apache.spark.storage.StorageLevel])在py4j.reflection.ReflectionEngine.getMethod(ReflectionEngine . java:335)py4j.reflection.ReflectionEngine.getMethod(ReflectionEngine.java:344)py4j.Gateway.invoke(Gateway.java:252)py4j的py4j.commands.AbstractCommand.invokeMethod(AbstractCommand.java:133) . 在java.lang.Thread.run(Thread.java:745)的py4j.GatewayConnection.run(GatewayConnection.java:209)上的commands.CallCommand.execute(CallCommand.java:79)

Spark 1.6.1

1 回答

  • 0

    端口号应为整数值 .

    items = stream.socketTextStream('localhost',9009)

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