Question

当我运行pyspark.SparkContext('loc', 'pyspark_rec')时，出现错误，说它无法解析master URL。作为spark编程的初学者，我不太清楚这意味着什么。但就我的代码而言，我没有使用任何部署模块（YARN，Hadoop等），而是在独立模式下测试代码。所以将URL指定为'loc'我相信很好。但有人可以向我解释我应该如何解决这个问题？谢谢。

以下是错误代码。

File "recommender.py", line 112, in spark_recommendations
    sc = pyspark.SparkContext('loc', 'pyspark_rec')



File "/Users/chlee021690/Desktop/Programming/spark/python/pyspark/context.py", line 134, in __init__
    self._jsc = self._initialize_context(self._conf._jconf)

  File "/Users/chlee021690/Desktop/Programming/spark/python/pyspark/context.py", line 180, in _initialize_context
    return self._jvm.JavaSparkContext(jconf)

  File "/Users/chlee021690/anaconda/lib/python2.7/site-packages/py4j/java_gateway.py", line 701, in __call__
    self._fqn)

  File "/Users/chlee021690/anaconda/lib/python2.7/site-packages/py4j/protocol.py", line 300, in get_return_value
    format(target_id, '.', name), value)

Py4JJavaError: An error occurred while calling None.org.apache.spark.api.java.JavaSparkContext.
: org.apache.spark.SparkException: Could not parse Master URL: 'loc'
    at org.apache.spark.SparkContext$.org$apache$spark$SparkContext$$createTaskScheduler(SparkContext.scala:1564)
    at org.apache.spark.SparkContext.<init>(SparkContext.scala:307)
    at org.apache.spark.api.java.JavaSparkContext.<init>(JavaSparkContext.scala:53)
    at sun.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method)
    at sun.reflect.NativeConstructorAccessorImpl.newInstance(NativeConstructorAccessorImpl.java:57)
    at sun.reflect.DelegatingConstructorAccessorImpl.newInstance(DelegatingConstructorAccessorImpl.java:45)
    at java.lang.reflect.Constructor.newInstance(Constructor.java:526)
    at py4j.reflection.MethodInvoker.invoke(MethodInvoker.java:234)
    at py4j.reflection.ReflectionEngine.invoke(ReflectionEngine.java:379)
    at py4j.Gateway.invoke(Gateway.java:214)
    at py4j.commands.ConstructorCommand.invokeConstructor(ConstructorCommand.java:79)
    at py4j.commands.ConstructorCommand.execute(ConstructorCommand.java:68)
    at py4j.GatewayConnection.run(GatewayConnection.java:207)
    at java.lang.Thread.run(Thread.java:744)

Answer 1

你会使用像

这样的东西

./bin/pyspark --master local[8]


from pyspark import SparkContext
sc = SparkContext("local", "context")

Answer 2

主网址通常是独立系统的IP地址（如果是服务器）或localhost。

独立模式：spark：// localhost：7077

服务器模式：spark：// your-master-server-ip-address：7077

希望这会有所帮助。干杯，阿希什

pyspark中的主URL是什么？

2 个答案: