当获取列使用的数组项时，可以拆分 when 如果元素为空，则更改为null：

// n is the max array size from split (in your example it's 3)
val n = 3

val df1 = df.withColumn(
    "ASplit",
    split(col("A"), "[.]")
  ).select(
    Seq(col("A")) ++ (0 to n-1).map(i =>
      when(col("ASplit")(i) === "", lit(null)).otherwise(col("ASplit")(i)).as(s"A split $i")
    ): _*
  )

//+-----+---------+---------+---------+
//|    A|A split 0|A split 1|A split 2|
//+-----+---------+---------+---------+
//|1.2.3|        1|        2|        3|
//| 4..5|        4|     null|        5|
//+-----+---------+---------+---------+

你可以 transform 通过将空值替换为空值来获得拆分结果：

val result = df.withColumn(
    "split",
    expr("transform(split(A, '\\\\.'), x -> case when x = '' then null else x end)")
).select($"A", $"split"(0), $"split"(1), $"split"(2))

result.show
+-----+--------+--------+--------+
|    A|split[0]|split[1]|split[2]|
+-----+--------+--------+--------+
|1.2.3|       1|       2|       3|
| 4..5|       4|    null|       5|
+-----+--------+--------+--------+