Question

我是Scala的新手。我正在尝试将scala列表（将源数据帧上的某些计算数据的结果保存）转换为Dataframe或Dataset。我没有找到任何直接的方法来做到这一点。但是，我尝试了以下过程将我的列表转换为DataSet，但它似乎无法正常工作。我提供以下3种情况。

有人可以给我一些希望，如何进行这种转换？感谢。

import org.apache.spark.sql.{DataFrame, Row, SQLContext, DataFrameReader}
import java.sql.{Connection, DriverManager, ResultSet, Timestamp}
import scala.collection._

case class TestPerson(name: String, age: Long, salary: Double)
var tom = new TestPerson("Tom Hanks",37,35.5)
var sam = new TestPerson("Sam Smith",40,40.5)

val PersonList = mutable.MutableList[TestPerson]()

//Adding data in list
PersonList += tom
PersonList += sam

//Situation 1: Trying to create dataset from List of objects:- Result:Error
//Throwing error
var personDS = Seq(PersonList).toDS()
/*
ERROR:
error: Unable to find encoder for type stored in a Dataset.  Primitive types
   (Int, String, etc) and Product types (case classes) are supported by     
importing sqlContext.implicits._  Support for serializing other types will  
be added in future releases.
     var personDS = Seq(PersonList).toDS()

*/
//Situation 2: Trying to add data 1-by-1 :- Result: not working as desired.    
the last record overwriting any existing data in the DS
var personDS = Seq(tom).toDS()
personDS = Seq(sam).toDS()

personDS += sam //not working. throwing error


//Situation 3: Working. However, I am having consolidated data in the list    
which I want to convert to DS; if I loop the results of the list in comma  
separated values and then pass that here, it will work but will create an  
extra loop in the code, which I want to avoid.
var personDS = Seq(tom,sam).toDS()
scala> personDS.show()
+---------+---+------+
|     name|age|salary|
+---------+---+------+
|Tom Hanks| 37|  35.5|
|Sam Smith| 40|  40.5|
+---------+---+------+

Answer 1

尝试不使用Seq：

case class TestPerson(name: String, age: Long, salary: Double)
val tom = TestPerson("Tom Hanks",37,35.5)
val sam = TestPerson("Sam Smith",40,40.5)
val PersonList = mutable.MutableList[TestPerson]()
PersonList += tom
PersonList += sam

val personDS = PersonList.toDS()
println(personDS.getClass)
personDS.show()

val personDF = PersonList.toDF()
println(personDF.getClass)
personDF.show()
personDF.select("name", "age").show()

<强>输出：

class org.apache.spark.sql.Dataset

+---------+---+------+
|     name|age|salary|
+---------+---+------+
|Tom Hanks| 37|  35.5|
|Sam Smith| 40|  40.5|
+---------+---+------+

class org.apache.spark.sql.DataFrame

+---------+---+------+
|     name|age|salary|
+---------+---+------+
|Tom Hanks| 37|  35.5|
|Sam Smith| 40|  40.5|
+---------+---+------+

+---------+---+
|     name|age|
+---------+---+
|Tom Hanks| 37|
|Sam Smith| 40|
+---------+---+

此外，请务必移动案例类TestPerson outside the scope of your object的声明。

Answer 2

使用序列：

val spark = SparkSession.builder().appName("Spark-SQL").master("local[2]").getOrCreate()

import spark.implicits._

var tom = new TestPerson("Tom Hanks",37,35.5)
var sam = new TestPerson("Sam Smith",40,40.5)

val PersonList = mutable.MutableList[TestPerson]()

//Adding data in list
PersonList += tom
PersonList += sam

//It will be work.
var personDS = Seq(PersonList).toDS()

SQLContext.implicits

Answer 3

case class TestPerson(name: String, age: Long, salary: Double)

val spark = SparkSession.builder().appName("List to Dataset").master("local[*]").getOrCreate()

var tom = new TestPerson("Tom Hanks",37,35.5)
var sam = new TestPerson("Sam Smith",40,40.5)
   
// mutable.MutableList[TestPerson]() is not required , i used below way which was 
// cleaner
val PersonList =  List(tom,sam)

import spark.implicits._
PersonList.toDS().show

将scala列表转换为DataFrame或DataSet

3 个答案: