如何将Spark列的名称作为字符串获取?

不确定这个问题是否真的得到了回答。您的函数可以这样实现( toString 返回列的名称):

def roundKeepname(c:Column,scale:Int) = round(c,scale).as(c.toString)

如果您不喜欢依赖toString,这里有一个更健壮的版本。您可以依赖于基础表达式,将其强制转换为NamedExpression并取其名称。

import org.apache.spark.sql.catalyst.expressions.NamedExpression
def roundKeepname(c:Column,scale:Int) = 
    c.expr.asInstanceOf[NamedExpression].name

它的工作原理是:

scala> spark.range(2).select(roundKeepname('id, 2)).show
+---+
| id|
+---+
|  0|
|  1|
+---+  

最后,如果您可以使用列的名称而不是列对象,那么您可以更改函数的签名,从而得到更简单的实现:

def roundKeepName(columnName:String, scale:Int) = 
    round(col(columnName),scale).as(columnName)

更新:

scala> val df = Seq((1.22,4.34,8.93),(3.44,12.66,17.44),(5.66,9.35,6.54)).toDF("x","y","z")
df: org.apache.spark.sql.DataFrame = [x: double, y: double ... 1 more field]

scala> df.show
+----+-----+-----+
|   x|    y|    z|
+----+-----+-----+
|1.22| 4.34| 8.93|
|3.44|12.66|17.44|
|5.66| 9.35| 6.54|
+----+-----+-----+


scala>  df.columns.foldLeft(df)( (acc,p)  => (acc.withColumn(p+"_t",round(col(p),1)).drop(p).withColumnRenamed(p+"_t",p))).show
+---+----+----+
|  x|   y|   z|
+---+----+----+
|1.2| 4.3| 8.9|
|3.4|12.7|17.4|
|5.7| 9.4| 6.5|
+---+----+----+


scala>