CODEWITHSUNDEEP
pythonapache-sparkpyspark
WestCoastProjects
WestCoastProjects

Reputation: 63062

How to access SparkContext in pyspark script

The following SOF question How to run script in Pyspark and drop into IPython shell when done? tells how to launch a pyspark script:

 %run -d myscript.py

But how do we access the existin spark context?

Just creating a new one does not work:

 ---->  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

But trying to use an existing one .. well what existing one?

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'>)

i.e. there is no variable for a SparkContext instance

Upvotes: 28

Views: 47205

Answers (4)

TechnoIndifferent
TechnoIndifferent

Reputation: 1114

Include the following:

from pyspark.context import SparkContext

and then invoke a static method on SparkContext as:

sc = SparkContext.getOrCreate()

Upvotes: 80

Rene B.
Rene B.

Reputation: 7364

If you created a already a SparkSession:

spark = SparkSession \
    .builder \
    .appName("StreamKafka_Test") \
    .getOrCreate()

Then you can access the "existing" SparkContext like this:

sc = spark.sparkContext

Upvotes: 10

vijay kumar
vijay kumar

Reputation: 2049

Standalone python script for wordcount : write a reusable spark context by using contextmanager

"""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"

to launch:

/bin/spark-submit SimpleApp.py

Upvotes: 4

mnm
mnm

Reputation: 2022

When you type pyspark at the terminal, python automatically creates the spark context sc.

Upvotes: 1

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