Scala Spark Dataframe Creation from Seq of Tuples: The Great Scala 2 vs Scala 3 Conundrum

Hey there, fellow Spark enthusiasts! Are you scratching your head over why creating a Spark Dataframe from a sequence of tuples works like a charm in Scala 2, but fails miserably in Scala 3? Well, you’re not alone! In this article, we’ll dive into the reasons behind this mysterious phenomenon and provide you with a step-by-step guide on how to overcome this hurdle in Scala 3.

The Problem: Scala Spark Dataframe Creation Fails in Scala 3

Let’s start with a simple example that works beautifully in Scala 2:

import org.apache.spark.sql.SparkSession

object DataFrameFromSeq {
  def main(args: Array[String]): Unit = {
    val spark = SparkSession.builder.appName("DataFrameFromSeq").getOrCreate()

    val data = Seq(("John", 25), ("Mary", 31), ("David", 42))
    val df = spark.createDataFrame(data)
    df.show()
  }
}

This code snippet creates a SparkSession, defines a sequence of tuples, and converts it into a Dataframe using the createDataFrame method. Running this code in Scala 2 yields the expected output:

_1	_2
John	25
Mary	31
David	42

However, when you try to run the same code in Scala 3, you’ll encounter a compilation error:

error: No implicit argument of type Encoder[(String, Int)] was found.
    val df = spark.createDataFrame(data)

What’s Causing the Issue: Scala 3’s Stricter Type Inference

The root cause of this problem lies in Scala 3’s more aggressive type inference, which is designed to reduce the need for explicit type annotations. While this new behavior is beneficial in many cases, it can sometimes lead to unexpected compilation errors.

In our example, the createDataFrame method requires an implicit Encoder[(String, Int)] to convert the sequence of tuples into a Dataframe. Scala 2’s type inference is lenient enough to automatically generate this implicit Encoder. However, Scala 3’s stricter type inference requires an explicit type annotation or a more precise type definition.

Solution 1: Providing an Explicit Encoder

One way to overcome this issue is to provide an explicit Encoder for the tuple type. You can do this by importing the necessary implicits and creating an instance of the Encoder[(String, Int)]:

import org.apache.spark.sql.Encoders

object DataFrameFromSeq {
  def main(args: Array[String]): Unit = {
    val spark = SparkSession.builder.appName("DataFrameFromSeq").getOrCreate()

    val data = Seq(("John", 25), ("Mary", 31), ("David", 42))
    implicit val encoder: Encoder[(String, Int)] = Encoders.tuple(Encoders.STRING, Encoders.INT)
    val df = spark.createDataFrame(data)
    df.show()
  }
}

This solution works because we’re explicitly defining the Encoder for the tuple type, which Scala 3 can then use to convert the sequence into a Dataframe.

Solution 2: Using Case Classes

Another approach is to define a case class to represent the tuple type and use it to create the sequence:

case class Person(name: String, age: Int)

object DataFrameFromSeq {
  def main(args: Array[String]): Unit = {
    val spark = SparkSession.builder.appName("DataFrameFromSeq").getOrCreate()

    val data = Seq(Person("John", 25), Person("Mary", 31), Person("David", 42))
    val df = spark.createDataFrame(data)
    df.show()
  }
}

In this example, we define a case class Person with two fields: name and age. We then create a sequence of Person instances, which can be converted into a Dataframe using the createDataFrame method.

Solution 3: Using a Custom Encoder

If you need more control over the encoding process, you can define a custom Encoder for the tuple type:

import org.apache.spark.sql.catalyst.encoders.RowEncoder
import org.apache.spark.sql.types.StructType

object DataFrameFromSeq {
  def main(args: Array[String]): Unit = {
    val spark = SparkSession.builder.appName("DataFrameFromSeq").getOrCreate()

    val data = Seq(("John", 25), ("Mary", 31), ("David", 42))
    val schema = new StructType().add("name", "string").add("age", "int")
    implicit val encoder: Encoder[(String, Int)] = RowEncoder(schema)
    val df = spark.createDataFrame(data)
    df.show()
  }
}

In this solution, we define a custom Encoder by creating a StructType schema that matches the tuple type and using the RowEncoder to convert the sequence into a Dataframe.

Conclusion

In this article, we explored the mysterious case of Scala Spark Dataframe creation from a sequence of tuples not working in Scala 3, but working in Scala 2. We delved into the reasons behind this issue, which is caused by Scala 3’s stricter type inference. We then presented three solutions to overcome this problem: providing an explicit Encoder, using case classes, and defining a custom Encoder.

By following these solutions, you should be able to create Spark Dataframes from sequences of tuples in Scala 3 without any issues. Remember to choose the solution that best fits your specific use case, and happy Spark-ing!

If you have any questions or need further assistance, feel free to ask in the comments below. Don’t forget to share this article with your fellow Scala enthusiasts to help them overcome this common hurdle.

References

• Apache Spark SQL Programming Guide
• Scala 3 Release Notes
• Scala 3 New Types

Happy coding!

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