下面是一个计算聚合的函数：

def compute(spark: SparkSession,
                  mydf: DataFrame): DataFrame = {
    
  mydf
    .groupBy(col("col1"), col("col2")
    .agg(
      count(when(col("col5") === "some_string_to_check", col("purchase_date"))).as("name1"),
      count(when(col("col5") === "some_string_to_check", col("purchase_date"))).as("name2"),
      count(when(col("col5") === "some_string_to_check"), col("purchase_date"))).as("name3"),
      count(when(col("col5") === "some_string_to_check"), col("purchase_date"))).as("name4"),

      count(when(col("col5") === "some_string1", col("purchase_date"))).as("name10"),
      count(when(col("col5") === "some_string1", col("purchase_date"))).as("name11"),
      count(when(col("col5") === "some_string1"), col("purchase_date"))).as("name12"),
      count(when(col("col5") === "some_string1"), col("purchase_date"))).as("name13")
    )
}

如您所见，此函数具有重复模式（我显示了2个，但实际上有10多个），因为这是重复代码，所以我希望将公共逻辑提取到函数中。类似于（伪代码）：

def compute(spark: SparkSession,
                  mydf: DataFrame): DataFrame = {
    
  mydf
    .groupBy(col("col1"), col("col2")
    .agg(
      func("col5", "some_string_to_check", "purchase_date", ["name1", "name2", "name3", "name4"]),
      func("col5", "some_string1", "purchase_date", ["name10", "name11", "name12", "name13"])
    )
}

def func(col, string_to_compate, date_col, array_of_name_results) = {
  count(col === string_to_compate, col(date_col))).as(array_of_name_results[0]),
  count(when(col  === string_to_compate, col(date_col))).as(array_of_name_results[1]),
  count(when(col  === string_to_compate), col(date_col))).as(array_of_name_results[2]),
  count(when(col === string_to_compate), col(date_col))).as(array_of_name_results[3]),
}

这可以做到吗？我找不到任何参考，如果它是可能的使用函数来创建.agg()的逻辑？任何帮助在这方面是感激。