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I have a dataframe df of columns ("id", "current_date", "days") and I am trying to add the the "days" to "current_date" and create a new dataframe with new column called "new_date" using spark scala function date_add()
val newDF = df.withColumn("new_Date", date_add(df("current_date"), df("days").cast("Int")))
But looks like the function date_add only accepts Int values and not columns. How can get the desired output in such case? Are there any alternative functions i can use to get the desired output?
spark version: 1.6.0
scala version: 2.10.6
No need to use an UDF, you can do it using an SQL expression:
val newDF = df.withColumn("new_date", expr("date_add(current_date,days)"))
A small custom udf can be used to make this date arithmetic possible.
import org.apache.spark.sql.functions.udf
import java.util.concurrent.TimeUnit
import java.util.Date
import java.text.SimpleDateFormat
val date_add = udf((x: String, y: Int) => {
val sdf = new SimpleDateFormat("yyyy-MM-dd")
val result = new Date(sdf.parse(x).getTime() + TimeUnit.DAYS.toMillis(y))
sdf.format(result)
} )
Usage:
scala> val df = Seq((1, "2017-01-01", 10), (2, "2017-01-01", 20)).toDF("id", "current_date", "days")
df: org.apache.spark.sql.DataFrame = [id: int, current_date: string, days: int]
scala> df.withColumn("new_Date", date_add($"current_date", $"days")).show()
+---+------------+----+----------+
| id|current_date|days| new_Date|
+---+------------+----+----------+
| 1| 2017-01-01| 10|2017-01-11|
| 2| 2017-01-01| 20|2017-01-21|
+---+------------+----+----------+
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
I have an instance of a RowMatrix that contains a single column.
I am trying to turn this RowMatrix into a dataframe but I am not quite sure how to convert org.apache.spark.rdd.RDD[org.apache.spark.mllib.linalg.Vector] to a dataframe.
val mat: RowMatrix = new RowMatrix(centred)
val mat_rows = mat.rows // RDD[Vector]
I know that if I had two (or more) columns I could use the following piece of code
val mat_rdd = mat_rows.map(_.toArray).map{case Array(p0, p1) => (p0, p1)}
val df = sparkSession.createDataFrame(mat_rdd).toDF("f1", "f2")
The problem is that I have just a single column. If I modify the above code to create a single-column dataframe I get the following error:
overloaded method value createDataFrame with alternatives:
It works fine in Spark 2.0.0:
import org.apache.spark.mllib.linalg.Vectors
import org.apache.spark.mllib.linalg.distributed.RowMatrix
val rows = spark.sparkContext.parallelize(Array(Vectors.dense(1), Vectors.dense(1)))
val mat: RowMatrix = new RowMatrix(rows)
scala> mat.rows.map(_.toArray).map{ case Array(a) => a}.toDF("f1").show()
+---+
| f1|
+---+
|1.0|
|1.0|
+---+
I have a dataframe with two columns one of which (called dist) is a dense vector. How can I convert it back into an array column of integers.
+---+-----+
| id| dist|
+---+-----+
|1.0|[2.0]|
|2.0|[4.0]|
|3.0|[6.0]|
|4.0|[8.0]|
+---+-----+
I tried using several variants of the following udf but it returns a type mismatch error
val toInt4 = udf[Int, Vector]({ (a) => (a)})
val result = df.withColumn("dist", toDf4(df("dist"))).select("dist")
I struggled for a while to get the answer from #ThomasLuechtefeld working. But was running into this very frustrating error:
org.apache.spark.sql.AnalysisException: cannot resolve 'UDF(features_scaled)' due to data type mismatch: argument 1 requires vector type, however, '`features_scaled`' is of vector type.
Turns out I needed to import DenseVector from the ml package instead of the mllib package.
So this worked for me:
import org.apache.spark.ml.linalg.DenseVector
import org.apache.spark.sql.functions._
val vectorToColumn = udf{ (x:DenseVector, index: Int) => x(index) }
myDataframe.withColumn("clusters_scaled",vectorToColumn(col("features_scaled"),lit(0)))
Yes, the only difference is that first line. This should absolutely be a comment, but I don't have the reputation. Sorry!
I think it's easiest to do it by going to the RDD API and then back.
import org.apache.spark.mllib.linalg.DenseVector
import org.apache.spark.sql.DataFrame
import org.apache.spark.rdd.RDD
import sqlContext._
// The original data.
val input: DataFrame =
sc.parallelize(1 to 4)
.map(i => i.toDouble -> new DenseVector(Array(i.toDouble * 2)))
.toDF("id", "dist")
// Turn it into an RDD for manipulation.
val inputRDD: RDD[(Double, DenseVector)] =
input.map(row => row.getAs[Double]("id") -> row.getAs[DenseVector]("dist"))
// Change the DenseVector into an integer array.
val outputRDD: RDD[(Double, Array[Int])] =
inputRDD.mapValues(_.toArray.map(_.toInt))
// Go back to a DataFrame.
val output = outputRDD.toDF("id", "dist")
output.show
You get:
+---+----+
| id|dist|
+---+----+
|1.0| [2]|
|2.0| [4]|
|3.0| [6]|
|4.0| [8]|
+---+----+
In spark 2.0 you can do something like:
import org.apache.spark.mllib.linalg.DenseVector
import org.apache.spark.sql.functions.udf
val vectorHead = udf{ x:DenseVector => x(0) }
df.withColumn("firstValue", vectorHead(df("vectorColumn")))
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