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我对 Pyspark 很陌生。我正在使用 Anaconda 执行一段 Pyspark 代码。我正在尝试启动 SparkContext 。这会引发下面列出的错误。看起来这个错误在 Python 中很正常，但我无法在 Pyspark 的上下文中弄清楚。代码是这样的：

import findspark
findspark.init()
import pyspark
import random
import os
import sys
sc = pyspark.SparkContext(appName="Pi")

仅执行此操作会引发如下错误：

---------------------------------------------------------------------------
UnboundLocalError                         Traceback (most recent call last)
<ipython-input-6-3fc08890af27> in <module>()
----> 1 sc = pyspark.SparkContext(appName="Pi")

C:\opt\spark\spark-2.3.0-bin-hadoop2.7\python\pyspark\context.py in __init__(self, master, appName, sparkHome, pyFiles, environment, batchSize, serializer, conf, gateway, jsc, profiler_cls)
    113         """
    114         self._callsite = first_spark_call() or CallSite(None, None, None)
--> 115         SparkContext._ensure_initialized(self, gateway=gateway, conf=conf)
    116         try:
    117             self._do_init(master, appName, sparkHome, pyFiles, environment, batchSize, serializer,

C:\opt\spark\spark-2.3.0-bin-hadoop2.7\python\pyspark\context.py in _ensure_initialized(cls, instance, gateway, conf)
    278         with SparkContext._lock:
    279             if not SparkContext._gateway:
--> 280                 SparkContext._gateway = gateway or launch_gateway(conf)
    281                 SparkContext._jvm = SparkContext._gateway.jvm
    282 

C:\opt\spark\spark-2.3.0-bin-hadoop2.7\python\pyspark\java_gateway.py in launch_gateway(conf)
     58                 submit_args
     59             ])
---> 60         command = command + shlex.split(submit_args)
     61 
     62         # Start a socket that will be used by PythonGatewayServer to communicate its port to us

UnboundLocalError: local variable 'submit_args' referenced before assignment
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