How do I import a local python module when running a python script in Oozie?

Question

I have two python files - my_python_A.py and my_python_B.py. The first file references the second (from my_python_B import *).

I'm executing the first python file from a shell action in Oozie (i.e. the script is simply python my_python_A.py), and am receiving the following error:

Traceback (most recent call last):
  File "my_python_A.py", line 2, in <module>
    from my_python_B import *
ImportError: No module named my_python_B
Failing Oozie Launcher, Main class [org.apache.oozie.action.hadoop.ShellMain], exit code [1]

Both python files are located under the same directory in HDFS. How can I get this import statement to work?

usually the interpreter looks first in the current directory, then in your shell, and then in your default PATH. Which environment are you using ? And, are these files part of the same project? If they are, perhaps you are better off using an __init__ file — mugabits
– mugabits, Commented Apr 25, 2016 at 18:20
@jmugz3 - not sure this is as relevant when running on a cluster, since all of the files are stored on a distributed system (i.e. HDFS). — John
– John, Commented Apr 25, 2016 at 18:23
Gotcha. I'm not familiar with Oozie, but it sounds like your interpreter is not recognizing your module, so you could try to add your working directory to your shell path. — mugabits
– mugabits, Commented Apr 25, 2016 at 18:29
you could try something like import sys sys.path.append("/Users/path/to/file") — jkhadka
– jkhadka, Commented Apr 25, 2016 at 19:09

Nitin Kumar · Accepted Answer · 2016-04-26 12:31:01Z

7

I faced the same issue and the way I worked around this problem was by setting the environment variable PYTHONPATH to the current working directory inside the shell script before I execute my python code

export PYTHONPATH=`pwd`
python m_python_A.py

Make sure that in your shell action you have included all the required python modules inside the <file></file> tags. Assuming that you have a shell script called sample_script.sh (inside which you have the aforementioned commands) your workflow.xml file should look something like this

<workflow-app name="shellTest" xmlns="uri:oozie:workflow:0.4">
    <start to="shell-action"/>
    <action name="shell-action">
        <shell xmlns="uri:oozie:shell-action:0.2">
            <job-tracker>${jobTracker}</job-tracker>
            <name-node>${nameNode}</name-node>
            <configuration>                
                <property>
                    <name>oozie.launcher.mapred.job.queue.name</name>
                    <value>${launcherqueue}</value>
                </property>
                <property>
                    <name>mapred.job.queue.name</name>
                    <value>${mapredqueue}</value>
                </property>
            </configuration>
            <exec>sample_script.sh</exec>
            <file>${appPath}/sample_script.sh#sample_script.sh</file>
            <file>${appPath}/m_python_A.py#m_python_A.py</file>
            <file>${appPath}/m_python_B.py#m_python_B.py</file>
            <capture-output/>
        </shell>
        <ok to="end"/>
        <error to="shell-action-failed"/>
    </action>

    <kill name="shell-action-failed">
        <message>Shell action failed, error message[${wf:errorMessage(wf:lastErrorNode())}]</message>
    </kill>

    <end name="end" />

</workflow-app>

answered Apr 26, 2016 at 12:31

Nitin Kumar

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2 Comments

Milimetric Over a year ago

What is ${appPath} in your example?

Nitin Kumar Over a year ago

${appPath} is the path to the folder on hdfs inside which your scripts lie.

ozw1z5rd · Accepted Answer · 2016-08-02 13:05:22Z

1

What about to add

sys.path.append(os.path.join(os.path.dirname(__file__), "lib"))

in your m_python_A.py to access anything which is stored into (I.E.) lib/ ?

answered Aug 2, 2016 at 13:05

ozw1z5rd

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