Scala 将字符串转换为类型column Spark
代码: 我正试图用上面的函数从现有的原始DF创建一个新的已处理DF 代码: 错误:Scala 将字符串转换为类型column Spark,scala,apache-spark,Scala,Apache Spark,代码: 我正试图用上面的函数从现有的原始DF创建一个新的已处理DF 代码: 错误: var processedDF = func(rawDF,"col1","col2") :73:错误:类型不匹配; 找到:字符串(“col1”) 必需:org.apache.spark.sql.Column var processedDF=func(rawDF,“col1”、“col2”) ^ 关于如何将函数参数的类型从String更改为org.apache.spark.sql.Column的任何建议 &l
var processedDF = func(rawDF,"col1","col2")
:73:错误:类型不匹配;
找到:字符串(“col1”)
必需:org.apache.spark.sql.Column
var processedDF=func(rawDF,“col1”、“col2”)
^
关于如何将函数参数的类型从String更改为org.apache.spark.sql.Column的任何建议
<console>:73: error: type mismatch;
found : String("col1")
required: org.apache.spark.sql.Column
var processedDF = func(rawDF,"col1","col2")
^
或
或者直接通过$
提供列
(其中spark
是SparkSession
对象)
或符号
import spark.implicits.StringToColumn
func(rawDF, $"col1", $"col2")
import org.apache.spark.sql.functions.col
func(rawDF, col("col1"), col("col2"))
func(rawDF, rawDF("col1"), rawDF("col2"))
import spark.implicits.StringToColumn
func(rawDF, $"col1", $"col2")
import spark.implicits.symbolToColumn
func(rawDF, 'col1, 'col2)