Question

我正在执行以下代码以从文本文件创建数据框。

    import org.apache.spark.SparkContext
import org.apache.spark.SparkConf
import org.apache.spark.sql.{SQLContext, Row}
import org.apache.spark.sql.types.{StructType, StringType, StructField}


/**
  * Created by PSwain on 6/19/2016.
  */
object RddToDataframe extends App {

  val scnf=new SparkConf().setAppName("RddToDataFrame").setMaster("local[1]")
  val sc = new SparkContext(scnf)
  val sqlContext = new SQLContext(sc)

  val employeeRdd=sc.textFile("C:\\Users\\pswain\\IdeaProjects\\test1\\src\\main\\resources\\employee")

  //Creating schema

  val employeeSchemaString="id name age"
  val schema = StructType(employeeSchemaString.split(",").map( colNmae => StructField(colNmae,StringType,true)))

  //Creating  RowRdd
  val rowRdd= employeeRdd.map(row => row.split(",")).map(row => Row(row(0).trim.toInt,row(1),row(2).trim.toInt))

  //Creating dataframe = RDD[rowRdd] + schema
  val employeeDF=sqlContext.createDataFrame(rowRdd,schema). registerTempTable("Employee")

  sqlContext.sql("select * from Employee").show()


}

但是在InteliJ中执行时，我发现类型不匹配错误如下所示。无法识别出现此错误的原因我只是将字符串转换为整数。员工文件具有以下输入，它们显示在一行中，但它们各为一行。

1201，饱和，25岁 1202，克里希纳，28岁 1203，amith，39 1204，javed，23 1205，prudvi，23

16/06/19 15:18:58 ERROR Executor: Exception in task 0.0 in stage 0.0 (TID 0)
scala.MatchError: 1201 (of class java.lang.Integer)
    at org.apache.spark.sql.catalyst.CatalystTypeConverters$StringConverter$.toCatalystImpl(CatalystTypeConverters.scala:295)
    at org.apache.spark.sql.catalyst.CatalystTypeConverters$StringConverter$.toCatalystImpl(CatalystTypeConverters.scala:294)
    at org.apache.spark.sql.catalyst.CatalystTypeConverters$CatalystTypeConverter.toCatalyst(CatalystTypeConverters.scala:102)

Answer 1

创建模式时，所有列类型都定义为StringType。

val schema = StructType(employeeSchemaString.split(",").map( colNmae => StructField(colNmae,StringType,true)))

但是rowRDD的类型为int，string和int。

这是工作代码

val structType= {
    val id = StructField("id", IntegerType)
    val name = StructField("name", StringType)
    val age = StructField("age", IntegerType)
    new StructType(Array(id, name , age))
}

val rowRdd= employeeRdd.map(row => row.split(",")).map(row => Row(row(0).trim().toInt,row(1),row(2).trim().toInt))

sqlContext.createDataFrame(rowRdd,structType). registerTempTable("Employee")

sqlContext.sql("select * from Employee").show()

面对＆＃34; scala.MatchError：1201（类java.lang.Integer）＆＃34;在创建数据框架时

1 个答案: