Scala Spark: Performance issue renaming huge number of columns - scala

To be able to work with columnnames of my DataFrame without escaping the . I need a function to "validify" all columnnames - but none of the methods I tried does the job in a timely manner (I'm aborting after 5 minutes).
The dataset I'm trying my algorithms on is the golub Dataset (get it here). It's a 2.2MB CSV file with 7200 columns. Renaming all columns should be a matter of seconds
Code to read the CSV in
var dfGolub = spark.read
.option("header", "true")
.option("inferSchema", "true")
.csv("golub_merged.csv")
.drop("_c0") // drop the first column
.repartition(numOfCores)
Attempts to rename columns:
def validifyColumnnames1(df : DataFrame) : DataFrame = {
import org.apache.spark.sql.functions.col
val cols = df.columns
val colsRenamed = cols.map(name => col(name).as(name.replaceAll("\\.","")))
df.select(colsRenamed : _*)
}
def validifyColumnnames2[T](df : Dataset[T]) : DataFrame = {
val newColumnNames = ArrayBuffer[String]()
for(oldCol <- df.columns) {
newColumnNames += oldCol.replaceAll("\\.","")
}
df.toDF(newColumnNames : _*)
}
def validifyColumnnames3(df : DataFrame) : DataFrame = {
var newDf = df
for(col <- df.columns){
newDf = newDf.withColumnRenamed(col,col.replaceAll("\\.",""))
}
newDf
}
Any ideas what is causing this performance issue?
Setup: I'm running Spark 2.1.0 on Ubuntu 16.04 in local[24] mode on a machine with 16cores * 2 threads and 96GB of RAM

Assuming you know the types you can simply create the schema instead of infering it (infering the schema costs performance and might even be wrong for csv).
Lets assume for simplicity you have the file example.csv as follows:
A.B, A.C, A.D
a,3,1
You can do something like this:
val scehma = StructType(Seq(StructField("A_B",StringType),StructField("A_C", IntegerType), StructField("AD", IntegerType)))
val df = spark.read.option("header","true").schema(scehma).csv("example.csv")
df.show()
+---+---+---+
|A_B|A_C| AD|
+---+---+---+
| a| 3| 1|
+---+---+---+
IF you do not know the info in advance you can use infer schema as you did before, then you can use the dataframe to generate the schema:
val fields = for {
x <- df.schema
} yield StructField(x.name.replaceAll("\\.",""), x.dataType, x.nullable)
val schema = StructType(fields)
and the reread the dataframe using that schema as before

Related

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

Dropping constant columns in a csv file

i would like to drop columns which are constant in a dataframe , here what i did , but i see that it tooks some much time to do it , special while writing the dataframe into the csv file , please any help to optimize the code to take less time
val spark = SparkSession.builder.master("local").appName("my-spark-app").getOrCreate()
val df = spark.read.option("inferSchema", "true").option("header", "false").csv("D:\\ProcessDataSet\\anis_data\\Set _1Mud Pumps_Merged.csv")
val aggregations = df.drop("DateTime").columns.map(c => stddev("c").as(c))
val df2 = df.agg(aggregations.head, aggregations.tail: _*)
val columnsToKeep: Seq[String] = (df2.first match {
case r : Row => r.toSeq.toArray.map(_.asInstanceOf[Double])
}).zip(df.columns)
.filter(_._1 != 0) // your special condition is in the filter
.map(_._2) // keep just the name of the column
// select columns with stddev != 0
val finalResult = df.select(columnsToKeep.head, columnsToKeep.tail : _*)
finalResult.write.option("header",true).csv("D:\\ProcessDataSet\\dataWithoutConstant\\Set _1Mud Pumps_MergedCleaned.csv")
}
I think there is no much room left for optimization. You are doing the right thing.
Maybe what you can try is to cache() your dataframe df.
df is used in two separate Spark actions so it is loaded twice.
Try :
...
val df = spark.read.option("inferSchema", "true").option("header", "false").csv("D:\\ProcessDataSet\\anis_data\\Set _1Mud Pumps_Merged.csv")
df.cache()
val aggregations = df.drop("DateTime").columns.map(c => stddev("c").as(c))
...

how many ways are there to add a new column to a data frame RDD in Spark API?

I can think of one only using withColumn():
val df = sc.dataFrame.withColumn('newcolname',{ lambda row: row + 1 } )
but how would I generalize this to Text data? For example of my DataFrame had
strning values say "This is an example of a string" and I wanted to extract the
first and last word as in val arraystring : Array[String] = Array(first,last)
Is this the thing you're looking for?
val sc: SparkContext = ...
val sqlContext = new SQLContext(sc)
import sqlContext.implicits._
val extractFirstWord = udf((sentence: String) => sentence.split(" ").head)
val extractLastWord = udf((sentence: String) => sentence.split(" ").reverse.head)
val sentences = sc.parallelize(Seq("This is an example", "And this is another one", "One_word", "")).toDF("sentence")
val splits = sentences
.withColumn("first_word", extractFirstWord(col("sentence")))
.withColumn("last_word", extractLastWord(col("sentence")))
splits.show()
Then the output is:
+--------------------+----------+---------+
| sentence|first_word|last_word|
+--------------------+----------+---------+
| This is an example| This| example|
|And this is anoth...| And| one|
| One_word| One_word| One_word|
| | | |
+--------------------+----------+---------+
# Create a simple DataFrame, stored into a partition directory
df1 = sqlContext.createDataFrame(sc.parallelize(range(1, 6))\
.map(lambda i: Row(single=i, double=i * 2)))
df1.save("data/test_table/key=1", "parquet")
# Create another DataFrame in a new partition directory,
# adding a new column and dropping an existing column
df2 = sqlContext.createDataFrame(sc.parallelize(range(6, 11))
.map(lambda i: Row(single=i, triple=i * 3)))
df2.save("data/test_table/key=2", "parquet")
# Read the partitioned table
df3 = sqlContext.parquetFile("data/test_table")
df3.printSchema()
https://spark.apache.org/docs/1.3.1/sql-programming-guide.html

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