Reputation: 1805
When I submit a python script as jar to spark action in oozie, I see the below error :
Traceback (most recent call last):
File "/home/hadoop/spark.py", line 5, in <module>
from pyspark import SparkContext, SparkConf
ImportError: No module named pyspark
Intercepting System.exit(1)
Although I can see that the pyspark libraries exist on my local FS :
$ ls /usr/lib/spark/python/pyspark/
accumulators.py heapq3.py rdd.py statcounter.py
broadcast.py __init__.py rddsampler.py status.py
cloudpickle.py java_gateway.py resultiterable.py storagelevel.py
conf.py join.py serializers.py streaming/
context.py ml/ shell.py tests.py
daemon.py mllib/ shuffle.py traceback_utils.py
files.py profiler.py sql/ worker.py
I know that there were issues with running pyspark on oozie like https://issues.apache.org/jira/browse/OOZIE-2482 but the error I am seeing is different from the JIRA ticket.
Also I am passing --conf spark.yarn.appMasterEnv.SPARK_HOME=/usr/lib/spark --conf spark.executorEnv.SPARK_HOME=/usr/lib/spark
as spark-opts
in my workflow definition.
Here is my sample application for reference :
masterNode ip-xxx-xx-xx-xx.ec2.internal
nameNode hdfs://${masterNode}:8020
jobTracker ${masterNode}:8032
master yarn
mode client
queueName default
oozie.libpath ${nameNode}/user/oozie/share/lib
oozie.use.system.libpath true
oozie.wf.application.path /user/oozie/apps/
<workflow-app name="spark-wf" xmlns="uri:oozie:workflow:0.5">
<start to="spark-action-test"/>
<action name="spark-action-test">
<spark xmlns="uri:oozie:spark-action:0.1">
<job-tracker>${jobTracker}</job-tracker>
<name-node>${nameNode}</name-node>
<configuration>
<property>
<name>mapred.compress.map.output</name>
<value>true</value>
</property>
</configuration>
<master>${master}</master>
<mode>${mode}</mode>
<name>Spark Example</name>
<jar>/home/hadoop/spark.py</jar>
<spark-opts>--driver-memory 512m --executor-memory 512m --num-executors 4 --conf spark.yarn.appMasterEnv.SPARK_HOME=/usr/lib/spark --conf spark.executorEnv.SPARK_HOME=/usr/lib/spark --conf spark.yarn.appMasterEnv.PYSPARK_PYTHON=/usr/lib/spark/python --conf spark.executorEnv.PYTHONPATH=/usr/lib/spark/python --files ${nameNode}/user/oozie/apps/hive-site.xml</spark-opts>
</spark>
<ok to="end"/>
<error to="kill"/>
</action>
<kill name="kill">
<message>Action failed, error message[${wf:errorMessage(wf:lastErrorNode())}]</message>
</kill>
<end name="end"/>
</workflow-app>
# sc is an existing SparkContext.
from pyspark import SparkContext, SparkConf
from pyspark.sql import HiveContext
conf = SparkConf().setAppName('test_pyspark_oozie')
sc = SparkContext(conf=conf)
sqlContext = HiveContext(sc)
sqlContext.sql("CREATE TABLE IF NOT EXISTS src (key INT, value STRING)")
As recommended here - http://www.learn4master.com/big-data/pyspark/run-pyspark-on-oozie, I also did put the following two files: py4j-0.9-src.zip pyspark.zip, under my ${nameNode}/user/oozie/share/lib folder.
I am using a single-node YARN cluster (AWS EMR) & trying to find out I can pass these pyspark modules to python in my oozie application. Any help is appreciated.
Upvotes: 0
Views: 2963
Reputation: 98
You are getting No module named error
because you have not mentioned PYTHONPATH
in your configuration. Add one more line in --conf
with PYTHONPATH=/usr/lib/spark/python
. I don't know how to set this PYTHONPATH
in oozie workflow defination but by adding PYTHONPATH
property in your configuration will definitely solve your problem.
Upvotes: 2