Question

我有一个如下所示的数据框

| id| age|   rbc|  bgr| dm|cad|appet| pe|ane|classification|
+---+----+------+-----+---+---+-----+---+---+--------------+
|  3|48.0|normal|117.0| no| no| poor|yes|yes|           ckd|
....
....
....

我编写了一个UDF，将分类yes, no, poor, normal转换为二进制0s和1s

def stringToBinary(stringValue: String): Int = {
    stringValue match {
        case "yes" => return 1
        case "no" => return 0
        case "present" => return 1
        case "notpresent" => return 0
        case "normal" => return 1
        case "abnormal" => return 0
    }
}

val stringToBinaryUDF = udf(stringToBinary _)

我将此应用于数据框，如下所示

val newCol = stringToBinaryUDF.apply(col("pc")) //creates the new column with formatted value
val refined1 = noZeroDF.withColumn("dm", newCol) //adds the new column to original

如何将多个列传递到UDF中，以便我不必为其他分类列重复自己？

Answer 1

如果udf函数执行与spark函数相同的工作，那么

udf函数不应该是选择，函数会对列数据进行序列化和反序列化。

将dataframe视为

+---+----+------+-----+---+---+-----+---+---+--------------+
|id |age |rbc   |bgr  |dm |cad|appet|pe |ane|classification|
+---+----+------+-----+---+---+-----+---+---+--------------+
|3  |48.0|normal|117.0|no |no |poor |yes|yes|ckd           |
+---+----+------+-----+---+---+-----+---+---+--------------+

您可以使用when函数

来达到您的要求

import org.apache.spark.sql.functions._
def applyFunction(column : Column) = when(column === "yes" || column === "present" || column === "normal", lit(1))
  .otherwise(when(column === "no" || column === "notpresent" || column === "abnormal", lit(0)).otherwise(column))

df.withColumn("dm", applyFunction(col("dm")))
  .withColumn("cad", applyFunction(col("cad")))
  .withColumn("rbc", applyFunction(col("rbc")))
  .withColumn("pe", applyFunction(col("pe")))
  .withColumn("ane", applyFunction(col("ane")))
  .show(false)

结果是

+---+----+---+-----+---+---+-----+---+---+--------------+
|id |age |rbc|bgr  |dm |cad|appet|pe |ane|classification|
+---+----+---+-----+---+---+-----+---+---+--------------+
|3  |48.0|1  |117.0|0  |0  |poor |1  |1  |ckd           |
+---+----+---+-----+---+---+-----+---+---+--------------+

现在问题清楚地表明，您不想重复所有列的过程，您可以执行以下操作

val columnsTomap = df.select("rbc", "cad", "rbc", "pe", "ane").columns

var tempdf = df
columnsTomap.map(column => {
  tempdf = tempdf.withColumn(column, applyFunction(col(column)))
})

tempdf.show(false)

Answer 2

UDF 可以使用许多参数，即许多列，但它应返回一个结果，即一列。

为了做到这一点，只需在stringToBinary功能中添加参数即可。

您希望它采用两列，如下所示：

def stringToBinary(stringValue: String, secondValue: String): Int = {
stringValue match {
    case "yes" => return 1
    case "no" => return 0
    case "present" => return 1
    case "notpresent" => return 0
    case "normal" => return 1
    case "abnormal" => return 0
}
}
val stringToBinaryUDF = udf(stringToBinary _)

希望这有帮助

Answer 3

您还可以使用foldLeft函数。将您的 UDF 称为stringToBinaryUDF：

import org.apache.spark.sql.functions._

val categoricalColumns = Seq("rbc", "cad", "rbc", "pe", "ane")
val refinedDF = categoricalColumns
    .foldLeft(noZeroDF) { (accumulatorDF: DataFrame, columnName: String) =>
         accumulatorDF
            .withColumn(columnName, stringToBinaryUDF(col(columnName)))
     }

这将尊重不变性和功能性编程。

将UDF应用于Spark Dataframe中的多个列

3 个答案: