I've got a unstructured RDD with keys and values. The values is of RDD[Any] and the keys are currently Strings, RDD[String] and mainly contain Maps. I would like to make them of type Row so I can make a dataframe eventually. Here is my rdd :
removed
Most of the rdd follows a pattern except for the last 4 keys, how should this be dealt with ? Perhaps split them into their own rdd, especially for reverseDeltas ?
Thanks
Edit
This is what I've tired so far based on the first answer below.
case class MyData(`type`: List[String], libVersion: Double, id: BigInt)
object MyDataBuilder{
def apply(s: Any): MyData = {
// read the input data and convert that to the case class
s match {
case Array(x: List[String], y: Double, z: BigInt) => MyData(x, y, z)
case Array(a: BigInt, Array(x: List[String], y: Double, z: BigInt)) => MyData(x, y, z)
case _ => null
}
}
}
val parsedRdd: RDD[MyData] = rdd.map(x => MyDataBuilder(x))
how it doesn't see to match any of those cases, how can I match on Map in scala ? I keep getting nulls back when printing out parsedRdd
To convert the RDD to a dataframe you need to have fixed schema. If you define the schema for the RDD rest is simple.
something like
val rdd2:RDD[Array[String]] = rdd.map( x => getParsedRow(x))
val rddFinal:RDD[Row] = rdd2.map(x => Row.fromSeq(x))
Alternate
case class MyData(....) // all the fields of the Schema I want
object MyDataBuilder {
def apply(s:Any):MyData ={
// read the input data and convert that to the case class
}
}
val rddFinal:RDD[MyData] = rdd.map(x => MyDataBuilder(x))
import spark.implicits._
val myDF = rddFinal.toDF
there is a method for converting an rdd to dataframe
use it like below
val rdd = sc.textFile("/pathtologfile/logfile.txt")
val df = rdd.toDF()
no you have dataframe do what ever you want on it using sql queries like below
val textFile = sc.textFile("hdfs://...")
// Creates a DataFrame having a single column named "line"
val df = textFile.toDF("line")
val errors = df.filter(col("line").like("%ERROR%"))
// Counts all the errors
errors.count()
// Counts errors mentioning MySQL
errors.filter(col("line").like("%MySQL%")).count()
// Fetches the MySQL errors as an array of strings
errors.filter(col("line").like("%MySQL%")).collect()
Related
I am trying to transform RDD[String] to RDD[Picture] but could not do it. If I could manage to convert RDD to RDD[Picture] I would use the def hasValidCountry to check if the values latitude and longitude of the picture meta valid. And after that I am trying to check if user Tags are valid with def hasTags in Picture class. The problem I encounter :
Implicit conversion found: row ⇒ augmentString(row): scala.collection.immutable.StringOps
type mismatch; found : String required: Array[String]
value InterestingPics is not a member of Array[Nothing] possible cause: maybe a semicolon is missing before `value InterestingPics'?
My intention is to choose line which has valid country and tags and transform all the line to new RDD[Picture] class.
ScalaFile1 (I have updated the ScalaFile):
object Part2 {
def main(args: Array[String]): Unit = {
var spark: SparkSession = null
try {
spark = SparkSession.builder().appName("Flickr using dataframes").config("spark.master", "local[*]").getOrCreate()
val originalFlickrMeta: RDD[String] = spark.sparkContext.textFile("flickrSample.txt")
val InterestingPics = originalFlickrMeta.map(row => row.split('\t')).map(field => Picture(field(0).toString())
InterestingPics.collect
InterestingPics.take(5).foreach(println)
This works, as an example:
case class case_for_rdd(c1: Int, c2: String, c3: String)
val rdd_data = spark.sparkContext.textFile("/FileStore/tables/csv01-4.txt")
val rdd = rdd_data.map(row => row.split(',')).map(field => case_for_rdd(field(0).toInt, field(1), field(2)))
rdd.collect
More complicated example with reading into RDD from file with array. Array needs a delimiter.
1,10,100,aa|bb|cc
2,20,200,xxxxxx|yyyyyyyy|z|aaa
Some sample code, but use List as otherwise you get to see array addresses, that's what those strange strings are, courtesy of smarter
people here:
case class case_for_rdd(c1: Int, c2: String, c3: String, a4: List[String])
val rdd_data = spark.sparkContext.textFile("/FileStore/tables/csv03.txt")
val myCaseRdd = rdd_data.map(row => row.split(',')).map(field => case_for_rdd(field(0).toInt, field(1), field(2), (field(3).split("\\|").toList)))
myCaseRdd.collect
My advice is to use a DF and the splitting stuff is then easier. Also, manipulation of the rdd via transformation, then the case class is lost. Array with DF api has no such issue.
I have an solution to my question in accordence with help of #thebluephantom. Thank you very much.
val InterestingPics = originalFlickrMeta.map(line => (new Picture(line.split("\t")))).filter(f => f.c != null && f.userTags.length > 0)
InterestingPics.collect().foreach(println)
Given a Breeze SparseVector object:
scala> val sv = new SparseVector[Double](Array(0, 4, 5), Array(1.5, 3.6, 0.4), 8)
sv: breeze.linalg.SparseVector[Double] = SparseVector(8)((0,1.5), (4,3.6), (5,0.4))
What is the best way to take the log of the values + 1?
Here is one way that works:
scala> new SparseVector(sv.index, log(sv.data.map(_ + 1)), sv.length)
res11: breeze.linalg.SparseVector[Double] = SparseVector(8)((0,0.9162907318741551), (4,1.5260563034950492), (5,0.3364722366212129))
I don't like this because it doesn't really make use of breeze to do the addition. We are using a breeze UFunc to take the log of an Array[Double], but that isn't much. I am concerned that in a distributed application with large SparseVectors, this will be slow.
Spark 1.6.3
You can define some UDF's to do arbitrary vectorized addition in Spark. First, you need to set up the ability to convert Spark vectors to Breeze vectors; an example of doing that is here. Once you have the implicit conversions in place, you have a few options.
To add any two columns you can use:
def addVectors(v1Col: String, v2Col: String, outputCol: String): DataFrame => DataFrame = {
// Error checking column names here
df: DataFrame => {
def add(v1: SparkVector, v2: SparkVector): SparkVector =
(v1.asBreeze + v2.asBreeze).fromBreeze
val func = udf((v1: SparkVector, v2: SparkVector) => add(v1, v2))
df.withColumn(outputCol, func(col(v1Col), col(v2Col)))
}
}
Note, the use of asBreeze and fromBreeze (as well as the alias for SparkVector) is established in the question linked above. A possible solution is to make a literal integer column by
df.withColumn(colName, lit(1))
and then add the columns.
The alternative for more complex mathematical functions is:
def applyMath(func: BreezeVector[Double] => BreezeVector[Double],
inColName: String, outColName: String): DataFrame => DataFrame = {
df: DataFrame => df.withColumn(outColName,
udf((v1: SparkVector) => func(v1.asBreeze).fromBreeze).apply(col(inColName)))
}
You could also make this generic in the Breeze vector parameter.
I want to parse the date columns in a DataFrame, and for each date column, the resolution for the date may change (i.e. 2011/01/10 => 2011 /01 if the resolution is set to "Month").
I wrote the following code:
def convertDataFrame(dataframe: DataFrame, schema : Array[FieldDataType], resolution: Array[DateResolutionType]) : DataFrame =
{
import org.apache.spark.sql.functions._
val convertDateFunc = udf{(x:String, resolution: DateResolutionType) => SparkDateTimeConverter.convertDate(x, resolution)}
val convertDateTimeFunc = udf{(x:String, resolution: DateResolutionType) => SparkDateTimeConverter.convertDateTime(x, resolution)}
val allColNames = dataframe.columns
val allCols = allColNames.map(name => dataframe.col(name))
val mappedCols =
{
for(i <- allCols.indices) yield
{
schema(i) match
{
case FieldDataType.Date => convertDateFunc(allCols(i), resolution(i)))
case FieldDataType.DateTime => convertDateTimeFunc(allCols(i), resolution(i))
case _ => allCols(i)
}
}
}
dataframe.select(mappedCols:_*)
}}
However it doesn't work. It seems that I can only pass Columns to UDFs. And I wonder if it will be very slow if I convert the DataFrame to RDD and apply the function on each row.
Does anyone know the correct solution? Thank you!
Just use a little bit of currying:
def convertDateFunc(resolution: DateResolutionType) = udf((x:String) =>
SparkDateTimeConverter.convertDate(x, resolution))
and use it as follows:
case FieldDataType.Date => convertDateFunc(resolution(i))(allCols(i))
On a side note you should take a look at sql.functions.trunc and sql.functions.date_format. These should at least part of the job without using UDFs at all.
Note:
In Spark 2.2 or later you can use typedLit function:
import org.apache.spark.sql.functions.typedLit
which support a wider range of literals like Seq or Map.
You can create a literal Column to pass to a udf using the lit(...) function defined in org.apache.spark.sql.functions
For example:
val takeRight = udf((s: String, i: Int) => s.takeRight(i))
df.select(takeRight($"stringCol", lit(1)))
I'm using the Cloudera's SparkOnHBase module in order to get data from HBase.
I get a RDD in this way:
var getRdd = hbaseContext.hbaseRDD("kbdp:detalle_feedback", scan)
Based on that, what I get is an object of type
RDD[(Array[Byte], List[(Array[Byte], Array[Byte], Array[Byte])])]
which corresponds to row key and a list of values. All of them represented by a byte array.
If I save the getRDD to a file, what I see is:
([B#f7e2590,[([B#22d418e2,[B#12adaf4b,[B#48cf6e81), ([B#2a5ffc7f,[B#3ba0b95,[B#2b4e651c), ([B#27d0277a,[B#52cfcf01,[B#491f7520), ([B#3042ad61,[B#6984d407,[B#f7c4db0), ([B#29d065c1,[B#30c87759,[B#39138d14), ([B#32933952,[B#5f98506e,[B#8c896ca), ([B#2923ac47,[B#65037e6a,[B#486094f5), ([B#3cd385f2,[B#62fef210,[B#4fc62b36), ([B#5b3f0f24,[B#8fb3349,[B#23e4023a), ([B#4e4e403e,[B#735bce9b,[B#10595d48), ([B#5afb2a5a,[B#1f99a960,[B#213eedd5), ([B#2a704c00,[B#328da9c4,[B#72849cc9), ([B#60518adb,[B#9736144,[B#75f6bc34)])
for each record (rowKey and the columns)
But what I need is to get the String representation of all and each of the keys and values. Or at least the values. In order to save it to a file and see something like
key1,(value1,value2...)
or something like
key1,value1,value2...
I'm completely new on spark and scala and it's being quite hard to get something.
Could you please help me with that?
First lets create some sample data:
scala> val d = List( ("ab" -> List(("qw", "er", "ty")) ), ("cd" -> List(("ac", "bn", "afad")) ) )
d: List[(String, List[(String, String, String)])] = List((ab,List((qw,er,ty))), (cd,List((ac,bn,afad))))
This is how the data is:
scala> d foreach println
(ab,List((qw,er,ty)))
(cd,List((ac,bn,afad)))
Convert it to Array[Byte] format
scala> val arrData = d.map { case (k,v) => k.getBytes() -> v.map { case (a,b,c) => (a.getBytes(), b.getBytes(), c.getBytes()) } }
arrData: List[(Array[Byte], List[(Array[Byte], Array[Byte], Array[Byte])])] = List((Array(97, 98),List((Array(113, 119),Array(101, 114),Array(116, 121)))), (Array(99, 100),List((Array(97, 99),Array(98, 110),Array(97, 102, 97, 100)))))
Create an RDD out of this data
scala> val rdd1 = sc.parallelize(arrData)
rdd1: org.apache.spark.rdd.RDD[(Array[Byte], List[(Array[Byte], Array[Byte], Array[Byte])])] = ParallelCollectionRDD[0] at parallelize at <console>:25
Create a conversion function from Array[Byte] to String:
scala> def b2s(a: Array[Byte]): String = new String(a)
b2s: (a: Array[Byte])String
Perform our final conversion:
scala> val rdd2 = rdd1.map { case (k,v) => b2s(k) -> v.map{ case (a,b,c) => (b2s(a), b2s(b), b2s(c)) } }
rdd2: org.apache.spark.rdd.RDD[(String, List[(String, String, String)])] = MapPartitionsRDD[1] at map at <console>:29
scala> rdd2.collect()
res2: Array[(String, List[(String, String, String)])] = Array((ab,List((qw,er,ty))), (cd,List((ac,bn,afad))))
I don't know about HBase but if those Array[Byte]s are Unicode strings, something like this should work:
rdd: RDD[(Array[Byte], List[(Array[Byte], Array[Byte], Array[Byte])])] = *whatever*
rdd.map(k, l =>
(new String(k),
l.map(a =>
a.map(elem =>
new String(elem)
)
))
)
Sorry for bad styling and whatnot, I am not even sure it will work.
I have two paired rdds in the form RDD [(String, mutable.HashSet[String]):
For example:
rdd1: 332101231222, "320758, 320762, 320760, 320759, 320757, 320761"
rdd2: 332101231222, "220758, 220762, 220760, 220759, 220757, 220761"
I want to combine rdd1 and rdd2 based on common keys, so o/p should be like:
332101231222 320758, 320762, 320760, 320759, 320757, 320761 220758, 220762, 220760, 220759, 220757, 220761
Here is my code:
def cogroupTest (rdd1: RDD [(String, mutable.HashSet[String])], rdd2: RDD [(String, mutable.HashSet[String])] ): Unit =
{
val prods_per_user_co_grouped = (rdd1).cogroup(rdd2)
prods_per_user_co_grouped.map { case (key: String, (value1: mutable.HashSet[String], value2: mutable.HashSet[String])) => {
val combinedhs = value1 ++ value2
val sstr = combinedhs.mkString("\t")
val keypadded = key + "\t"
s"$keypadded$sstr"
}
}.saveAsTextFile("/scratch/rdds_joined/")
Here is the error that I get when I run the my program:
scala.MatchError: (32101231222,(CompactBuffer(Set(320758, 320762, 320760, 320759, 320757, 320761)),CompactBuffer(Set(220758, 220762, 220760, 220759, 220757, 220761)))) (of class scala.Tuple2)
Any help with this will be great!
As you might guess from the name cogroup groups observations by key. It means that in your case you get:
(String, (Iterable[mutable.HashSet[String]], Iterable[mutable.HashSet[String]]))
not
(String, (mutable.HashSet[String], mutable.HashSet[String]))
It is pretty clear when you take a look at the error you get. If you want to combine pairs you should use join method. If not you should adjust pattern to match structure you get and then use something like this:
val combinedhs = value1.reduce(_ ++ _) ++ value2.reduce(_ ++ _)