How to convert rdd object to dataframe in spark

Question

How can I convert an RDD  org apache spark rdd RDD org apache spark sql Row   to a Dataframe org apache spark sql DataFrame  I converted a dataframe to rdd using  rdd  After processing it I want it back in dataframe  How can I do this

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One needs to create a schema  and attach it to the Rdd    Assuming val spark is a product of a SparkSession builder         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  gt          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    Same steps  but with fewer val declarations       val arrayOfArrayOfAnys Array          Array  aString  0L          3 14159           Array  bString  9876543210L 2 71828             val rddOfRows spark sparkContext parallelize arrayOfArrayOfAnys  map f  gt 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

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Method 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     Method 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    gt  temp val1 val3    toDF   rows show     Method 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     Method 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     Extracted the value from the row object and then applied the case class to convert rdd to DF  val temp1   attrib1 map case Row   key  Int     gt  s  key    val temp2   attrib2 map case Row   key  Int    gt  s  key     case class RLT  id  String  attrib 1   String  attrib 2   String  import hiveContext implicits    val df   result map  s   gt  RLT s 0  s 1  s 2     toDF

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Assuming your RDD row  is called rdd  you can use   val sqlContext   new SQLContext sc   import sqlContext implicits   rdd toDF

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Suppose you have a DataFrame and you want to do some modification on the fields data by converting it to RDD Row    val aRdd   aDF map x  gt Row x getAs Long   id   x getAs List String    role   head     To convert back to DataFrame from RDD we need to define the structure type of the RDD   If the datatype was Long  then it will become as LongType in structure   If String then StringType in structure   val aStruct   new StructType Array StructField  id  LongType nullable   true  StructField  role  StringType nullable   true      Now you can convert the RDD to DataFrame using the createDataFrame method   val aNamedDF   sqlContext createDataFrame aRdd aStruct

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SparkSession has a number of createDataFrame methods that create a DataFrame given an RDD  I imagine one of these will work for your context  For example  def createDataFrame rowRDD  RDD Row   schema  StructType   DataFrame   Creates a DataFrame from an RDD containing Rows using the given schema

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On newer versions of spark  2 0    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

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Note  This answer was originally posted here   I am posting this answer because I would like to share additional details about the available options that I did not find in the other answers     To create a DataFrame from an RDD of Rows  there are two main options   1  As already pointed out  you could use toDF   which can be imported by import sqlContext implicits    However  this approach only works for the following types of RDDs    RDD Int  RDD Long  RDD String  RDD T  lt   scala Product     source  Scaladoc of the SQLContext implicits object   The last signature actually means that it can work for an RDD of tuples or an RDD of case classes  because tuples and case classes are subclasses of scala Product    So  to use this approach for an RDD Row   you have to map it to an RDD T  lt   scala Product   This can be done by mapping each row to a custom case class or to a tuple  as in the following code snippets   val df   rdd map      case Row val1  String       valN  Long    gt   val1       valN     toDF  col1 name         colN name     or  case class MyClass val1  String       valN  Long   0L  val df   rdd map      case Row val1  String       valN  Long    gt  MyClass val1       valN     toDF  col1 name         colN name     The main drawback of this approach  in my opinion  is that you have to explicitly set the schema of the resulting DataFrame in the map function  column by column  Maybe this can be done programatically if you don t know the schema in advance  but things can get a little messy there  So  alternatively  there is another option     2  You can use createDataFrame rowRDD  RDD Row   schema  StructType  as in the accepted answer  which is available in the SQLContext object  Example for converting an RDD of an old DataFrame   val rdd   oldDF rdd val newDF   oldDF sqlContext createDataFrame rdd  oldDF schema    Note that there is no need to explicitly set any schema column  We reuse the old DF s schema  which is of StructType class and can be easily extended  However  this approach sometimes is not possible  and in some cases can be less efficient than the first one

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Here is a simple example of converting your List into Spark RDD and then converting that Spark RDD into Dataframe   Please note that I have used Spark-shell s scala REPL to execute following code  Here sc is an instance of SparkContext which is implicitly available in Spark-shell  Hope it answer your question   scala gt  val numList   List 1 2 3 4 5  numList  List Int    List 1  2  3  4  5   scala gt  val numRDD   sc parallelize numList  numRDD  org apache spark rdd RDD Int    ParallelCollectionRDD 80  at parallelize at  lt console gt  28  scala gt  val numDF   numRDD toDF numDF  org apache spark sql DataFrame     1  int   scala gt  numDF show  ---     1   ---     1     2     3     4     5   ---

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I tried to explain the solution using the word count problem   1  Read the file using sc    Produce word count Methods to create DF   rdd toDF method      rdd toDF  word   count      spark createDataFrame rdd schema        Read file using spark  val rdd sc textFile  D   cca175 data        Rdd to Dataframe  val df sc textFile  D   cca175 data    toDF  t1        df show  Method 1  Create word count RDD to Dataframe  val df rdd flatMap x  gt x split       map x  gt  x 1   reduceByKey  x y   gt  x y   toDF  word   count     Method2  Create Dataframe from Rdd  val df spark createDataFrame wordRdd     with header    val df spark createDataFrame wordRdd  toDF  word   count    df show   Method3  Define Schema  import org apache spark sql types     val schema new StructType                        add StructField  word  StringType true                        add StructField  count  StringType true    Create RowRDD  import org apache spark sql Row val rowRdd wordRdd map x  gt  Row x  1 x  2           Create DataFrame from RDD with schema  val df spark createDataFrame rowRdd schema  df show

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To convert an Array Row  to DataFrame or Dataset  the following works elegantly   Say  schema is the StructType for the row then  val rows  Array Row      implicit val encoder   RowEncoder apply schema  import spark implicits   rows toDS

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This code works perfectly from Spark 2 x with Scala 2 11  Import necessary classes  import org apache spark sql  Row  SparkSession  import org apache spark sql types  DoubleType  StringType  StructField  StructType    Create SparkSession Object  and Here it s spark   val spark  SparkSession   SparkSession builder master  local   getOrCreate val sc   spark sparkContext    Just used to create test RDDs   Let s an RDD to make it 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           Method 1  Using 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    ------ --------------------    Method 2  Using SparkSession createDataFrame RDD obj  and specifying column names    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    ------ --------------------    Method 3  Actual answer to the question   This way requires the input rdd should be of 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          create the schema   val schema   new StructType      add StructField  id   StringType  true      add StructField  val1   DoubleType  true      add StructField  val2   DoubleType  true     Now apply both rowsRdd and schema to 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   ------ ---- ----

[scala] How to convert rdd object to dataframe in spark

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