如何访问pyspark脚本中的SparkContext

Question

下面的SOF问题How to run script in Pyspark and drop into IPython shell when done?告诉您如何启动pyspark脚本：

 %run -d myscript.py

但是我们如何访问existin spark上下文呢？

仅仅创建一个新的不起作用：

 ---->  sc = SparkContext("local", 1)

 ValueError: Cannot run multiple SparkContexts at once; existing 
 SparkContext(app=PySparkShell, master=local) created by <module> at 
 /Library/Python/2.7/site-packages/IPython/utils/py3compat.py:204

但试图使用现有的。。什么是现有的？

In [50]: for s in filter(lambda x: 'SparkContext' in repr(x[1]) and len(repr(x[1])) < 150, locals().iteritems()):
    print s
('SparkContext', <class 'pyspark.context.SparkContext'>)

也就是说，SparkContext实例没有变量

Answer 1

wordcount的独立python脚本：使用contextmanager编写可重用的spark上下文

"""SimpleApp.py"""
from contextlib import contextmanager
from pyspark import SparkContext
from pyspark import SparkConf


SPARK_MASTER='local'
SPARK_APP_NAME='Word Count'
SPARK_EXECUTOR_MEMORY='200m'

@contextmanager
def spark_manager():
    conf = SparkConf().setMaster(SPARK_MASTER) \
                      .setAppName(SPARK_APP_NAME) \
                      .set("spark.executor.memory", SPARK_EXECUTOR_MEMORY)
    spark_context = SparkContext(conf=conf)

    try:
        yield spark_context
    finally:
        spark_context.stop()

with spark_manager() as context:
    File = "/home/ramisetty/sparkex/README.md"  # Should be some file on your system
    textFileRDD = context.textFile(File)
    wordCounts = textFileRDD.flatMap(lambda line: line.split()).map(lambda word: (word, 1)).reduceByKey(lambda a, b: a+b)
    wordCounts.saveAsTextFile("output")

print "WordCount - Done"

启动：

/bin/spark-submit SimpleApp.py

Answer 2

从pyspark.context导入SparkContext

然后在SparkContext上调用静态方法：

sc = SparkContext.getOrCreate()