filter after flatMap spark - scala

I have a simple example
val arrayStructureData = Seq(
Row("test1|value1"),
Row("test2|value2"),
Row("test3|value3")
)
val arrayStructureSchema = new StructType()
.add("name", StringType)
val df = spark.createDataFrame(
spark.sparkContext.parallelize(arrayStructureData), arrayStructureSchema)
import spark.implicits._
val distPhens = df.flatMap(row => row.getString(0).split("\\|"))
.filter(x => x.like("test[0-9]+"))
.toDF("distinct_phens")
where I'm trying to run filter after running flatMap. The desired output is :
value1
value2
value3
If I understand correctly, like expects a column but I am not sure how to "refer" to the column after flatMap has been executed.
I need this filter operation to run after flatMap.
Thanks in advance.


You can refer to the column object using col, and do an rlike filter:
val result = df.flatMap(row => row.getString(0).split("\\|")).filter(col("value").rlike("test[0-9]+"))
result.show
+-----+
|value|
+-----+
|test1|
|test2|
|test3|
+-----+

Related

what is the order guarantee when joining two columns of a spark dataframe which are processed separately?

I have dataframe with 3 columns
date
jsonString1
jsonString2
I want to expand attributes inside json into columns. so i did something like this.
val json1 = spark.read.json(dataframe.select(col("jsonString1")).rdd.map(_.getString(0)))
val json2 = spark.read.json(dataframe.select(col("jsonString2")).rdd.map(_.getString(0)))
val json1Table = json1.selectExpr("id", "status")
val json2Table = json2.selectExpr("name", "address")
now i want to put these table together. so i did the following
val json1TableWithIndex = addColumnIndex(json1Table)
val json2TableWithIndex = addColumnIndex(json2Table)
var finalResult = json1Table
.join(json2Table, Seq("columnindex"))
.drop("columnindex")
def addColumnIndex(df: DataFrame) = spark.createDataFrame(
df.rdd.zipWithIndex.map { case (row, columnindex) => Row.fromSeq(row.toSeq :+ columnindex) },
StructType(df.schema.fields :+ StructField("columnindex", LongType, false))
)
After sampling few rows I observe that rows match exactly as in the source dataframe
I did not find any information on the order guarantee when joining two columns of a dataframe which are processed separately. Is this the right way to solve my problem. Any help is appreciated.

It is always risky to rely on undocumented behaviours, as your code might not work as you intended, because you only have a partial understanding of it.
You can do the same thing in a much more efficient way without using any split and join approach. Use a from_json function to create two nested columns and then flatten out the nested columns and finally drop out the intermediate JSON string columns and nested columns.
Here is an example fo the whole process.
import org.apache.spark.sql.types.{StringType, StructType, StructField}
val df = (Seq(
("09-02-2020","{\"id\":\"01\", \"status\":\"Active\"}","{\"name\":\"Abdullah\", \"address\":\"Jumeirah\"}"),
("10-02-2020","{\"id\":\"02\", \"status\":\"Dormant\"}","{\"name\":\"Ali\", \"address\":\"Jebel Ali\"}")
).toDF("date","jsonString1","jsonString2"))
scala> df.show()
+----------+--------------------+--------------------+
| date| jsonString1| jsonString2|
+----------+--------------------+--------------------+
|09-02-2020|{"id":"01", "stat...|{"name":"Abdullah...|
|10-02-2020|{"id":"02", "stat...|{"name":"Ali", "a...|
+----------+--------------------+--------------------+
val schema1 = (StructType(Seq(
StructField("id", StringType, true),
StructField("status", StringType, true)
)))
val schema2 = (StructType(Seq(
StructField("name", StringType, true),
StructField("address", StringType, true)
)))
val dfFlattened = (df.withColumn("jsonData1", from_json(col("jsonString1"), schema1))
.withColumn("jsonData2", from_json(col("jsonString2"), schema2))
.withColumn("id", col("jsonData1.id"))
.withColumn("status", col("jsonData1.status"))
.withColumn("name", col("jsonData2.name"))
.withColumn("address", col("jsonData2.address"))
.drop("jsonString1")
.drop("jsonString2")
.drop("jsonData1")
.drop("jsonData2"))
scala> dfFlattened.show()
+----------+---+-------+--------+---------+
| date| id| status| name| address|
+----------+---+-------+--------+---------+
|09-02-2020| 01| Active|Abdullah| Jumeirah|
|10-02-2020| 02|Dormant| Ali|Jebel Ali|
+----------+---+-------+--------+---------+

Spark 2.3: subtract dataframes but preserve duplicate values (Scala)

Copying example from this question:
As a conceptual example, if I have two dataframes:
words = [the, quick, fox, a, brown, fox]
stopWords = [the, a]
then I want the output to be, in any order:
words - stopWords = [quick, brown, fox, fox]
ExceptAll can do this in 2.4 but I cannot upgrade. The answer in the linked question is specific to a dataframe:
words.join(stopwords, words("id") === stopwords("id"), "left_outer")
.where(stopwords("id").isNull)
.select(words("id")).show()
as in you need to know the pkey and the other columns.
Can anyone come up with an answer that will work on any dataframe?

Here is an implementation for you all. I have tested in Spark 2.4.2, it should work for 2.3 too (not 100% sure)
val df1 = spark.createDataset(Seq("the","quick","fox","a","brown","fox")).toDF("c1")
val df2 = spark.createDataset(Seq("the","a")).toDF("c1")
exceptAllCustom(df1, df2, Seq("c1")).show()
def exceptAllCustom(df1 : DataFrame, df2 : DataFrame, pks : Seq[String]): DataFrame = {
val notNullCondition = pks.foldLeft(lit(0==0))((column,cName) => column && df2(cName).isNull)
val joinCondition = pks.foldLeft(lit(0==0))((column,cName) => column && df2(cName)=== df1(cName))
val result = df1.join(df2, joinCondition, "left_outer")
.where(notNullCondition)
pks.foldLeft(result)((df,cName) => df.drop(df2(cName)))
}
Result -
+-----+
| c1|
+-----+
|quick|
| fox|
|brown|
| fox|
+-----+

Turns out it's easier to do df1.except(df2) and then join the results with df1 to get all the duplicates.
Full code:
def exceptAllCustom(df1: DataFrame, df2: DataFrame): DataFrame = {
val except = df1.except(df2)
val columns = df1.columns
val colExpr: Column = df1(columns.head) <=> except(columns.head)
val joinExpression = columns.tail.foldLeft(colExpr) { (colExpr, p) =>
colExpr && df1(p) <=> except(p)
}
val join = df1.join(except, joinExpression, "inner")
join.select(df1("*"))
}

Apply Function to DataFrame Rows and Convert Back to DataFrame Spark / Scala [duplicate]

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 ?

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

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.

Assuming your RDD[row] is called rdd, you can use:
val sqlContext = new SQLContext(sc)
import sqlContext.implicits._
rdd.toDF()

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 <: 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 <: 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) => (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) => 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.

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=>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)

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) => 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 ) => 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

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> 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|
+---+

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: _*)

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=>
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=>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

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=>x.split(" ")).map(x=>(x,1)).reduceByKey((x,y)=>(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=>(Row(x._1,x._2)))
Create DataFrame from RDD with schema
val df=spark.createDataFrame(rowRdd,schema)
df.show

I meet the same problem, and finally solve it. It's quite simple and easy.
You have to add this code import sc.implicits._, sc means SQLContext. add this code you will get rdd.toDF() method.
Transform your rdd[RawData] to rdd[YourCaseClass]. For example, you have a rdd type like this rdd[(String, Integer, Long)], you can create a Case Class YourCaseClass(name: String, age: Integer, timestamp: Long) and convert raw rdd to rdd with YourCaseClass type, then you get rdd[YourCaseClass]
save rdd[YourCaseClass] to hive table. yourRdd.toDF().write.format("parquet").mode(SaveMode.Overwrite).insertInto(yourHiveTableName) Use case class to represent rdd type, we can avoid naming each column field or StructType related schema.

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

Creating a Spark DataFrame from a single string

I'm trying to take a hardcoded String and turn it into a 1-row Spark DataFrame (with a single column of type StringType) such that:
String fizz = "buzz"
Would result with a DataFrame whose .show() method looks like:
+-----+
| fizz|
+-----+
| buzz|
+-----+
My best attempt thus far has been:
val rawData = List("fizz")
val df = sqlContext.sparkContext.parallelize(Seq(rawData)).toDF()
df.show()
But I get the following compiler error:
java.lang.ClassCastException: org.apache.spark.sql.types.ArrayType cannot be cast to org.apache.spark.sql.types.StructType
at org.apache.spark.sql.SQLContext.createDataFrame(SQLContext.scala:413)
at org.apache.spark.sql.SQLImplicits.rddToDataFrameHolder(SQLImplicits.scala:155)
Any ideas as to where I'm going awry? Also, how do I set "buzz" as the row value for the fizz column?
Update:
Trying:
sqlContext.sparkContext.parallelize(rawData).toDF()
I get a DF that looks like:
+----+
| _1|
+----+
|buzz|
+----+

Try:
sqlContext.sparkContext.parallelize(rawData).toDF()
In 2.0 you can:
import spark.implicits._
rawData.toDF
Optionally provide a sequence of names for toDF:
sqlContext.sparkContext.parallelize(rawData).toDF("fizz")

In Java, the following works:
List<String> textList = Collections.singletonList("yourString");
SQLContext sqlContext = new SQLContext(sparkContext);
Dataset<Row> data = sqlContext
.createDataset(textList, Encoders.STRING())
.withColumnRenamed("value", "text");

How to convert rdd object to dataframe in spark

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 ?

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

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.

Assuming your RDD[row] is called rdd, you can use:
val sqlContext = new SQLContext(sc)
import sqlContext.implicits._
rdd.toDF()

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 <: 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 <: 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) => (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) => 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.

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=>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)

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) => 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 ) => 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

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> 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|
+---+

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: _*)

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=>
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=>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

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=>x.split(" ")).map(x=>(x,1)).reduceByKey((x,y)=>(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=>(Row(x._1,x._2)))
Create DataFrame from RDD with schema
val df=spark.createDataFrame(rowRdd,schema)
df.show

I meet the same problem, and finally solve it. It's quite simple and easy.
You have to add this code import sc.implicits._, sc means SQLContext. add this code you will get rdd.toDF() method.
Transform your rdd[RawData] to rdd[YourCaseClass]. For example, you have a rdd type like this rdd[(String, Integer, Long)], you can create a Case Class YourCaseClass(name: String, age: Integer, timestamp: Long) and convert raw rdd to rdd with YourCaseClass type, then you get rdd[YourCaseClass]
save rdd[YourCaseClass] to hive table. yourRdd.toDF().write.format("parquet").mode(SaveMode.Overwrite).insertInto(yourHiveTableName) Use case class to represent rdd type, we can avoid naming each column field or StructType related schema.

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