Scala-Spark Dynamically call groupby and agg with parameter values

I want to write a custom grouping and aggregate function to get user specified column names and user specified aggregation map.I do not know the column names and aggregation map up front. I want to write a function similar to something like below. But i am new to Scala and i cannot solve it.

def groupAndAggregate(df: DataFrame,  aggregateFun: Map[String, String], cols: List[String] ): DataFrame ={
  val grouped = df.groupBy(cols)
  val aggregated = grouped.agg(aggregateFun)
  aggregated.show()
}

and want to call it like

val listOfStrings =  List("A", "B", "C")
val result = groupAndAggregate(df, Map("D"-> "SUM", "E"-> "COUNT"), listOfStrings)

How can i do this? Can anyone help me please.

Solution 1:

Your code is almost correct - with two issues:

1. The return type of your function is DataFrame, but the last line is aggregated.show(), which returns Unit. Remove the call to show to return aggregated itself, or just return the result of agg immediately

2. DataFrame.groupBy expects arguments as follows: col1: String, cols: String* - so you need to pass matching arguments: the first columns, and then the rest of the columns as a list of arguments, you can do that as follows: df.groupBy(cols.head, cols.tail: _*)

Altogether, your function would be:

def groupAndAggregate(df: DataFrame,  aggregateFun: Map[String, String], cols: List[String] ): DataFrame ={
  val grouped = df.groupBy(cols.head, cols.tail: _*)
  val aggregated = grouped.agg(aggregateFun)
  aggregated
}

Or, a similar shorter version:

def groupAndAggregate(df: DataFrame,  aggregateFun: Map[String, String], cols: List[String] ): DataFrame = {
  df.groupBy(cols.head, cols.tail: _*).agg(aggregateFun)
}

If you do want to call show within your function:

def groupAndAggregate(df: DataFrame,  aggregateFun: Map[String, String], cols: List[String] ): DataFrame ={
  val grouped = df.groupBy(cols.head, cols.tail: _*)
  val aggregated = grouped.agg(aggregateFun)
  aggregated.show()
  aggregated
}