将Scala列表转换为DataFrame或DataSet

Question

将Scala列表转换为DataFrame或DataSet

scalaapache-sparkapache-spark-sqlapache-spark-datasetapache-spark-encoders

10

我是Scala的新手。我正在尝试将Scala列表（其中包含源DataFrame上某些计算数据的结果）转换为Dataframe或Dataset。我没有找到任何直接的方法来完成这个操作。但是，我已经尝试了以下过程来将我的列表转换为DataSet，但似乎无法工作。下面提供了3种情况。

请问有人可以给我一些希望之光，告诉我如何进行这种转换吗？谢谢。

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|
+---------+---+------+

- Leo

你的Spark和Scala版本是什么？ - Ajeet Shah

Spark版本为1.6.1。 - Leo

3个回答

1

使用 Seq：

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

- zhming

1

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

- kartik kudada

网页内容由stack overflow 提供, 点击上面的

可以查看英文原文，
原文链接

- Ajeet Shah · Accepted Answer

不使用 Seq 进行尝试：

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()

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|
+---------+---+

另外，请确保将TestPerson案例类的声明移动到对象范围之外。

请注意，这是已经翻译好的文本，不需要再次翻译。