如何在火花中将rdd对象转换为数据帧


139

如何将RDD(org.apache.spark.rdd.RDD[org.apache.spark.sql.Row])转换为数据框org.apache.spark.sql.DataFrame。我使用将数据框转换为rdd .rdd。处理完后,我希望它返回到数据框中。我怎样才能做到这一点 ?


Answers:


88

SqlContext有许多createDataFrame创建DataFrame给定值的方法RDD。我想其中一种将适合您的情况。

例如:

def createDataFrame(rowRDD: RDD[Row], schema: StructType): DataFrame

使用给定架构从包含行的RDD创建一个DataFrame。


92

此代码可完美地在Spark 2.x和Scala 2.11中正常工作

导入必要的类

import org.apache.spark.sql.{Row, SparkSession}
import org.apache.spark.sql.types.{DoubleType, StringType, StructField, StructType}

创建SparkSession对象,在这里spark

val spark: SparkSession = SparkSession.builder.master("local").getOrCreate
val sc = spark.sparkContext // Just used to create test RDDs

让我们RDD来做吧DataFrame

val rdd = sc.parallelize(
  Seq(
    ("first", Array(2.0, 1.0, 2.1, 5.4)),
    ("test", Array(1.5, 0.5, 0.9, 3.7)),
    ("choose", Array(8.0, 2.9, 9.1, 2.5))
  )
)

方法一

使用SparkSession.createDataFrame(RDD obj)

val dfWithoutSchema = spark.createDataFrame(rdd)

dfWithoutSchema.show()
+------+--------------------+
|    _1|                  _2|
+------+--------------------+
| first|[2.0, 1.0, 2.1, 5.4]|
|  test|[1.5, 0.5, 0.9, 3.7]|
|choose|[8.0, 2.9, 9.1, 2.5]|
+------+--------------------+

方法2

使用SparkSession.createDataFrame(RDD obj)并指定列名。

val dfWithSchema = spark.createDataFrame(rdd).toDF("id", "vals")

dfWithSchema.show()
+------+--------------------+
|    id|                vals|
+------+--------------------+
| first|[2.0, 1.0, 2.1, 5.4]|
|  test|[1.5, 0.5, 0.9, 3.7]|
|choose|[8.0, 2.9, 9.1, 2.5]|
+------+--------------------+

方法3(问题的实际答案)

这种方式要求输入rdd应为type RDD[Row]

val rowsRdd: RDD[Row] = sc.parallelize(
  Seq(
    Row("first", 2.0, 7.0),
    Row("second", 3.5, 2.5),
    Row("third", 7.0, 5.9)
  )
)

创建模式

val schema = new StructType()
  .add(StructField("id", StringType, true))
  .add(StructField("val1", DoubleType, true))
  .add(StructField("val2", DoubleType, true))

现在同时将rowsRddschema应用于createDataFrame()

val df = spark.createDataFrame(rowsRdd, schema)

df.show()
+------+----+----+
|    id|val1|val2|
+------+----+----+
| first| 2.0| 7.0|
|second| 3.5| 2.5|
| third| 7.0| 5.9|
+------+----+----+

2
感谢您展示以一种可以理解的方式使用createDataFrame的不同方式
vatsug

第三种方法对数据砖很有用,因为其他砖无法正常工作并给出错误
Narendra Maru

67

假设您的RDD [row]被称为rdd,则可以使用:

val sqlContext = new SQLContext(sc) 
import sqlContext.implicits._
rdd.toDF()

26
我认为它不适用于RDD [Row]。我有什么想念的吗?
丹尼尔·德·保拉

4
由于Spark2.0将SQLContext替换为SparkSession,但该类保留在代码库中以实现向后兼容(scaladoc)。使用它会引发弃用警告。
tomaskazemekas

18

注意:此答案最初发布在这里

我发布此答案是因为我想分享有关其他答案中未找到的可用选项的更多详细信息


要从RDD行创建数据帧,有两个主要选项:

1)前面已经指出的那样,你可以使用toDF()它可以通过进口import sqlContext.implicits._。但是,此方法仅适用于以下类型的RDD:

  • RDD[Int]
  • RDD[Long]
  • RDD[String]
  • RDD[T <: scala.Product]

(来源:Scaladoc所述的SQLContext.implicits对象)

最后一个签名实际上意味着它可以用于元组的RDD或案例类的RDD(因为元组和案例类是的子类scala.Product)。

因此,要将这种方法用于RDD[Row],您必须将其映射到RDD[T <: scala.Product]。可以通过将每一行映射到自定义案例类或元组来完成,如以下代码段所示:

val df = rdd.map({ 
  case Row(val1: String, ..., valN: Long) => (val1, ..., valN)
}).toDF("col1_name", ..., "colN_name")

要么

case class MyClass(val1: String, ..., valN: Long = 0L)
val df = rdd.map({ 
  case Row(val1: String, ..., valN: Long) => MyClass(val1, ..., valN)
}).toDF("col1_name", ..., "colN_name")

这种方法的主要缺点(我认为)是,您必须在map函数中逐列显式设置结果DataFrame的架构。如果您事先不知道该模式,则可以通过编程方式完成,但是事情可能会变得有些混乱。因此,还有另一种选择:


2)您可以createDataFrame(rowRDD: RDD[Row], schema: StructType)在接受的答案中使用as,该答案在SQLContext对象中可用。转换旧DataFrame的RDD的示例:

val rdd = oldDF.rdd
val newDF = oldDF.sqlContext.createDataFrame(rdd, oldDF.schema)

请注意,无需显式设置任何架构列。我们重用了旧的DF模式,该模式属于StructType类并且可以轻松扩展。但是,这种方法有时是不可能的,并且在某些情况下可能不如第一种方法有效。


感谢您的详细信息import sqlContext.implicits.
-javadba

将来,请不要针对多个问题发布相同的答案。如果问题是重复的,则张贴一个好的答案,然后投票或举旗以将另一个问题作为重复的问题结束。如果问题不是重复的,请定制问题的答案。请参阅我如何写一个好的答案?

15

假设您有个,DataFrame并且想要通过将其转换为来对字段数据进行一些修改RDD[Row]

val aRdd = aDF.map(x=>Row(x.getAs[Long]("id"),x.getAs[List[String]]("role").head))

转换回DataFrameRDD我们需要定义结构类型RDD

如果是数据类型,Long 则它将变为LongType结构化。

如果StringStringType在结构上。

val aStruct = new StructType(Array(StructField("id",LongType,nullable = true),StructField("role",StringType,nullable = true)))

现在,您可以使用createDataFrame方法将RDD转换为DataFrame 。

val aNamedDF = sqlContext.createDataFrame(aRdd,aStruct)

7

这是将List转换为Spark RDD,然后将该Spark RDD转换为Dataframe的简单示例。

请注意,我已经使用Spark-shell的scala REPL执行以下代码,这里sc是SparkContext的实例,该实例在Spark-shell中隐式可用。希望它能回答您的问题。

scala> val numList = List(1,2,3,4,5)
numList: List[Int] = List(1, 2, 3, 4, 5)

scala> val numRDD = sc.parallelize(numList)
numRDD: org.apache.spark.rdd.RDD[Int] = ParallelCollectionRDD[80] at parallelize at <console>:28

scala> val numDF = numRDD.toDF
numDF: org.apache.spark.sql.DataFrame = [_1: int]

scala> numDF.show
+---+
| _1|
+---+
|  1|
|  2|
|  3|
|  4|
|  5|
+---+

一个有趣的事实:当列表为Double而不是int(或Long,String,<:Product)时,这将停止工作。
瑞克·莫里兹

不回答OP:谈论RDD [Row]
javadba

6

方法1:(Scala)

val sqlContext = new org.apache.spark.sql.SQLContext(sc)
import sqlContext.implicits._
val df_2 = sc.parallelize(Seq((1L, 3.0, "a"), (2L, -1.0, "b"), (3L, 0.0, "c"))).toDF("x", "y", "z")

方法2:(Scala)

case class temp(val1: String,val3 : Double) 

val rdd = sc.parallelize(Seq(
  Row("foo",  0.5), Row("bar",  0.0)
))
val rows = rdd.map({case Row(val1:String,val3:Double) => temp(val1,val3)}).toDF()
rows.show()

方法1:(Python)

from pyspark.sql import Row
l = [('Alice',2)]
Person = Row('name','age')
rdd = sc.parallelize(l)
person = rdd.map(lambda r:Person(*r))
df2 = sqlContext.createDataFrame(person)
df2.show()

方法2:(Python)

from pyspark.sql.types import * 
l = [('Alice',2)]
rdd = sc.parallelize(l)
schema =  StructType([StructField ("name" , StringType(), True) , 
StructField("age" , IntegerType(), True)]) 
df3 = sqlContext.createDataFrame(rdd, schema) 
df3.show()

从行对象中提取值,然后应用案例类将rdd转换为DF

val temp1 = attrib1.map{case Row ( key: Int ) => s"$key" }
val temp2 = attrib2.map{case Row ( key: Int) => s"$key" }

case class RLT (id: String, attrib_1 : String, attrib_2 : String)
import hiveContext.implicits._

val df = result.map{ s => RLT(s(0),s(1),s(2)) }.toDF

4

在较新版本的Spark(2.0+)上

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

val spark = SparkSession
  .builder()
  .getOrCreate()
import spark.implicits._

val dfSchema = Seq("col1", "col2", "col3")
rdd.toDF(dfSchema: _*)

1
sparkSession只是sqlContext,hiveContext的包装器
Archit

1
One needs to create a schema, and attach it to the Rdd.

假设val spark是SparkSession.builder的产品...

    import org.apache.spark._
    import org.apache.spark.sql._       
    import org.apache.spark.sql.types._

    /* Lets gin up some sample data:
     * As RDD's and dataframes can have columns of differing types, lets make our
     * sample data a three wide, two tall, rectangle of mixed types.
     * A column of Strings, a column of Longs, and a column of Doubules 
     */
    val arrayOfArrayOfAnys = Array.ofDim[Any](2,3)
    arrayOfArrayOfAnys(0)(0)="aString"
    arrayOfArrayOfAnys(0)(1)=0L
    arrayOfArrayOfAnys(0)(2)=3.14159
    arrayOfArrayOfAnys(1)(0)="bString"
    arrayOfArrayOfAnys(1)(1)=9876543210L
    arrayOfArrayOfAnys(1)(2)=2.71828

    /* The way to convert an anything which looks rectangular, 
     * (Array[Array[String]] or Array[Array[Any]] or Array[Row], ... ) into an RDD is to 
     * throw it into sparkContext.parallelize.
     * http://spark.apache.org/docs/latest/api/scala/index.html#org.apache.spark.SparkContext shows
     * the parallelize definition as 
     *     def parallelize[T](seq: Seq[T], numSlices: Int = defaultParallelism)
     * so in our case our ArrayOfArrayOfAnys is treated as a sequence of ArraysOfAnys.
     * Will leave the numSlices as the defaultParallelism, as I have no particular cause to change it. 
     */
    val rddOfArrayOfArrayOfAnys=spark.sparkContext.parallelize(arrayOfArrayOfAnys)

    /* We'll be using the sqlContext.createDataFrame to add a schema our RDD.
     * The RDD which goes into createDataFrame is an RDD[Row] which is not what we happen to have.
     * To convert anything one tall and several wide into a Row, one can use Row.fromSeq(thatThing.toSeq)
     * As we have an RDD[somethingWeDontWant], we can map each of the RDD rows into the desired Row type. 
     */     
    val rddOfRows=rddOfArrayOfArrayOfAnys.map(f=>
        Row.fromSeq(f.toSeq)
    )

    /* Now to construct our schema. This needs to be a StructType of 1 StructField per column in our dataframe.
     * https://spark.apache.org/docs/latest/api/scala/index.html#org.apache.spark.sql.types.StructField shows the definition as
     *   case class StructField(name: String, dataType: DataType, nullable: Boolean = true, metadata: Metadata = Metadata.empty)
     * Will leave the two default values in place for each of the columns:
     *        nullability as true, 
     *        metadata as an empty Map[String,Any]
     *   
     */

    val schema = StructType(
        StructField("colOfStrings", StringType) ::
        StructField("colOfLongs"  , LongType  ) ::
        StructField("colOfDoubles", DoubleType) ::
        Nil
    )

    val df=spark.sqlContext.createDataFrame(rddOfRows,schema)
    /*
     *      +------------+----------+------------+
     *      |colOfStrings|colOfLongs|colOfDoubles|
     *      +------------+----------+------------+
     *      |     aString|         0|     3.14159|
     *      |     bString|9876543210|     2.71828|
     *      +------------+----------+------------+
    */ 
    df.show 

步骤相同,但val声明较少:

    val arrayOfArrayOfAnys=Array(
        Array("aString",0L         ,3.14159),
        Array("bString",9876543210L,2.71828)
    )

    val rddOfRows=spark.sparkContext.parallelize(arrayOfArrayOfAnys).map(f=>Row.fromSeq(f.toSeq))

    /* If one knows the datatypes, for instance from JDBC queries as to RDBC column metadata:
     * Consider constructing the schema from an Array[StructField].  This would allow looping over 
     * the columns, with a match statement applying the appropriate sql datatypes as the second
     *  StructField arguments.   
     */
    val sf=new Array[StructField](3)
    sf(0)=StructField("colOfStrings",StringType)
    sf(1)=StructField("colOfLongs"  ,LongType  )
    sf(2)=StructField("colOfDoubles",DoubleType)        
    val df=spark.sqlContext.createDataFrame(rddOfRows,StructType(sf.toList))
    df.show

1

我试图使用 字数统计问题。1.使用sc读取文件

  1. 产生字数
  2. 创建DF的方法

    • rdd.toDF方法
    • rdd.toDF(“ word”,“ count”)
      • spark.createDataFrame(rdd,schema)

    使用Spark读取文件

    val rdd=sc.textFile("D://cca175/data/")  

    Rdd到数据框

    val df = sc.textFile(“ D:// cca175 / data /”).toDF(“ t1”)df.show

    方法一

    创建字数RDD到数据框

    val df=rdd.flatMap(x=>x.split(" ")).map(x=>(x,1)).reduceByKey((x,y)=>(x+y)).toDF("word","count")

    方法2

    从Rdd创建数据框

    val df=spark.createDataFrame(wordRdd) 
    # with header   
    val df=spark.createDataFrame(wordRdd).toDF("word","count")  df.show

    方法3

    定义架构

    导入org.apache.spark.sql.types._

    val schema = new StructType()。add(StructField(“ word”,StringType,true))。add(StructField(“ count”,StringType,true))

    创建RowRDD

    import org.apache.spark.sql.Row
    val rowRdd=wordRdd.map(x=>(Row(x._1,x._2)))     

    使用模式从RDD创建DataFrame

    val df = spark.createDataFrame(rowRdd,schema)
    df.show


0

要将Array [Row]转换为DataFrame或Dataset,可以使用以下方法:

假设模式是该行的StructType,然后

val rows: Array[Row]=...
implicit val encoder = RowEncoder.apply(schema)
import spark.implicits._
rows.toDS
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