两种方法都有效。
使用sql方法是可行的。不要使用

，这将包括旧列，只需
CONCAT
并重命名为
AS
.

customers.createOrReplaceTempView("customers")
spark.sql("SELECT CONCAT(name, ' ', last_name) AS name FROM customers").show()
//+--------+
//|    name|
//+--------+
//|John Doe|
//|Jane Doe|
//+--------+

withColumn
同样的，还有一个
withColumnRenamed
.
因此，按照您的意愿执行操作，创建一个新列，然后删除原始列并重命名新列。

// Problem Setup
val customers = = Seq(("John", "Doe"), ("Jane", "Doe")).toDF("name", "last_name")

customers.show()
//+----+---------+
//|name|last_name|
//+----+---------+
//|John|      Doe|
//|Jane|      Doe|
//+----+---------+

import org.apache.spark.sql.functions.{lit, col, concat}

customers.withColumn(
  "name_last_name", concat(col("name"), lit(" "), col("last_name"))
).drop("name", "last_name").withColumnRenamed("name_last_name", "name").show()
//+--------+
//|    name|
//+--------+
//|John Doe|
//|Jane Doe|
//+--------+

当然，你可以继续做手术本身在
withColumn
函数调用，为新生成的列提供标签
name
取代了旧的，但你还是得放弃
last_name
.

customers.withColumn(
  "name", concat(col("name"), lit(" "), col("last_name"))
).drop("last_name").show()
//+--------+
//|    name|
//+--------+
//|John Doe|
//|Jane Doe|
//+--------+