随着问题的发展，我将原来的答案留在了那里，这解决了最后一个问题.

As the question went through an evolution I leave the original answer there and this addresses the final question.

重要的一点，现在可以满足以下要求的输入内容:

Important point, the input mentioned as follows is now catered for:

val df0 = Seq (
            ("ABC", List(Map("C" -> "1", "D" -> "2"), Map("C" -> "3", "D" -> "4"))),
            ("XYZ", List(Map("C" -> "44", "D" -> "55"), Map("C" -> "188", "D" -> "199"), Map("C" -> "88", "D" -> "99")))
              )
             .toDF("A", "B")

也可以这样做，但是随后需要对此脚本进行修改，尽管这很简单:

Can also be done like this, but then the script needs to be modified for this, although trivial:

val df0 = Seq (
           ("ABC", List(Map("C" -> "1",  "D" -> "2"))), 
           ("ABC", List(Map("C" -> "44", "D" -> "55"))),
           ("XYZ", List(Map("C" -> "11", "D" -> "22")))
              )
            .toDF("A", "B")

然后按照要求的格式进行操作:

Following on from requested format then:

val df1 = df0.select($"A", explode($"B")).toDF("A", "Bn")

val df2 = df1.withColumn("SeqNum", monotonically_increasing_id()).toDF("A", "Bn", "SeqNum") 

val df3 = df2.select($"A", explode($"Bn"), $"SeqNum").toDF("A", "B", "C", "SeqNum")

val df4 = df3.withColumn("dummy", concat( $"SeqNum", lit("||"), $"A"))

val df5 = df4.select($"dummy", $"B", $"C").groupBy("dummy").pivot("B").agg(first($"C")) 

val df6 = df5.withColumn("A", substring_index(col("dummy"), "||", -1)).drop("dummy")

df6.show(false)

返回:

+---+---+---+
|C  |D  |A  |
+---+---+---+
|3  |4  |ABC|
|1  |2  |ABC|
|88 |99 |XYZ|
|188|199|XYZ|
|44 |55 |XYZ|
+---+---+---+

您可以重新排序列.