How to use a Scala class inside Pyspark

Yes it is possible although can be far from trivial. Typically you want a Java (friendly) wrapper so you don’t have to deal with Scala features which cannot be easily expressed using plain Java and as a result don’t play well with Py4J gateway.

Assuming your class is int the package com.example and have Python DataFrame called df

df = ... # Python DataFrame

you’ll have to:

1. Build a jar using your favorite build tool.

2. Include it in the driver classpath for example using --driver-class-path argument for PySpark shell / spark-submit. Depending on the exact code you may have to pass it using --jars as well

3. Extract JVM instance from a Python SparkContext instance:

   jvm = sc._jvm
   

4. Extract Scala SQLContext from a SQLContext instance:

   ssqlContext = sqlContext._ssql_ctx
   

5. Extract Java DataFrame from the df:

   jdf = df._jdf
   

6. Create new instance of SimpleClass:

   simpleObject = jvm.com.example.SimpleClass(ssqlContext, jdf, "v")
   

7. Callexe method and wrap the result using Python DataFrame:

   from pyspark.sql import DataFrame
   
   DataFrame(simpleObject.exe(), ssqlContext)
   

The result should be a valid PySpark DataFrame. You can of course combine all the steps into a single call.

Important: This approach is possible only if Python code is executed solely on the driver. It cannot be used inside Python action or transformation. See How to use Java/Scala function from an action or a transformation? for details.