I have a Seq of Tuples:
val seqTuple: Seq[(String, Future[String])] = Seq(("A", Future("X")), ("B", Future("Y")))
and I want to get:
val futureSeqTuple: Future[Seq[(String, String)]] = Future(Seq(("A", "X"), ("B", "Y")))
I know I can do:
val futureSeq: Future[Seq[String]] = Future.sequence(seqTuple.map(_._2))
but I am losing the first String in the Tuple.
What is the best way to get a Future[Seq[(String, String)]]?
Use the futures in tuples to map each tuple to future of tuple first,
then sequence:
Future.sequence(
seqTuple.map{case (s1, fut_s2) => fut_s2.map{s2 => (s1, s2)} }
)
Step by step, from inner terms to outer terms:
The inner map converts Future("X") to Future(("A", "X")).
The outer map converts each ("A", Future("X")) into an Future(("A", "X")), thus giving you a Seq[Future[(String, String)]].
Now you can use sequence on that to obtain Future[Seq[(String, String)]]
The answer given here works fine, but I think Future.traverse would work more succinctly here:
Future.traverse(seqTuple) {
case (s1, s2Future) => s2Future.map{ s2 => (s1, s2) }
}
This function involves converting the input argument :)
Related
I am trying to apply a function to cartesian RDDs. The function is taken from here and I have no idea how to make it work on cartesian RDDs.
val combined = rdd_valid.cartesian(rdd1)
combined.collect().foreach(a => println(a))
(abcde,abdce)
(somethin,somthing)
(afghr, decsvt)
My first thought was to do
val newRDD = combined.map(Levenshtein.distance)
But it doesn't work.
Assuming combined has the type RDD[(String, String)], and Levenshtein.distance has this signature:
def distance(s1:String, s2:String)
You can apply it as follows:
val newRDD = combined.map { case (s1, s2) => Levenshtein.distance(s1, s2) }
Or, alternatively:
val newRDD = combined.map(t => Levenshtein.distance(t._1, t._2))
I am using Spark 1.5.1 with Scala 2.10.5
I have an RDD[Array[String], Vector] for each element of the RDD:
I want to take each String in the Array[String] and combine it
with the Vector to create a tuple (String, Vector), this step will lead to the creation of several tuples from each element of the initial RDD
The goal is to end by building an RDD of tuples: RDD[(String,
Vector)], this RDD contains all the tuples created in the previous step.
Thanks
Consider this :
rdd.flatMap { case (arr, vec) => arr.map( (s) => (s, vec) ) }
(The first flatMap lets you get a RDD[(String, Vector)] as an output as opposed to a map which would get you a RDD[Array[(String, Vector)]])
Have you tried this?
// rdd: RDD[Array[String], Vector] - initial RDD
val new_rdd = rdd
.flatMap {
case (array: Array[String], vec: Vector) => array.map(str => (str, vec))
}
Toy example (I'm running it in spark-shell):
val rdd = sc.parallelize(Array((Array("foo", "bar"), 100), (Array("one", "two"), 200)))
val new_rdd = rdd
.map {
case (array: Array[String], vec: Int) => array.map(str => (str, vec))
}
.flatMap(arr => arr)
new_rdd.collect
res14: Array[(String, Int)] = Array((foo,100), (bar,100), (one,200), (two,200))
Imagine you have a Map[Option[Int], String] and you want to have a Map[Int, String] discarding the entry which contain None as the key.
Another example, that should be somehow similar is List[(Option[Int], String)] and transform it to List[(Int, String)], again discarding the tuple which contain None as the first element.
What's the best approach?
collect is your friend here:
example data definition
val data = Map(Some(1) -> "data", None -> "")
solution for Map
scala> data collect { case ( Some(i), s) => (i,s) }
res4: scala.collection.immutable.Map[Int,String] = Map(1 -> data)
the same approach works for a list of tuples
scala> data.toList collect { case ( Some(i), s) => (i,s) }
res5: List[(Int, String)] = List((1,data))
I have a list of Tuples of type : (user id, name, count).
For example,
val x = sc.parallelize(List(
("a", "b", 1),
("a", "b", 1),
("c", "b", 1),
("a", "d", 1))
)
I'm attempting to reduce this collection to a type where each
element name is counted.
So in above val x is converted to :
(a,ArrayBuffer((d,1), (b,2)))
(c,ArrayBuffer((b,1)))
Here is the code I am currently using :
val byKey = x.map({case (id,uri,count) => (id,uri)->count})
val grouped = byKey.groupByKey
val count = grouped.map{case ((id,uri),count) => ((id),(uri,count.sum))}
val grouped2: org.apache.spark.rdd.RDD[(String, Seq[(String, Int)])] = count.groupByKey
grouped2.foreach(println)
I'm attempting to use reduceByKey as it performs faster than groupByKey.
How can reduceByKey be implemented instead of above code to provide
the same mapping ?
Following your code:
val byKey = x.map({case (id,uri,count) => (id,uri)->count})
You could do:
val reducedByKey = byKey.reduceByKey(_ + _)
scala> reducedByKey.collect.foreach(println)
((a,d),1)
((a,b),2)
((c,b),1)
PairRDDFunctions[K,V].reduceByKey takes an associative reduce function that can be applied to the to type V of the RDD[(K,V)]. In other words, you need a function f[V](e1:V, e2:V) : V . In this particular case with sum on Ints: (x:Int, y:Int) => x+y or _ + _ in short underscore notation.
For the record: reduceByKey performs better than groupByKey because it attemps to apply the reduce function locally before the shuffle/reduce phase. groupByKey will force a shuffle of all elements before grouping.
Your origin data structure is: RDD[(String, String, Int)], and reduceByKey can only be used if data structure is RDD[(K, V)].
val kv = x.map(e => e._1 -> e._2 -> e._3) // kv is RDD[((String, String), Int)]
val reduced = kv.reduceByKey(_ + _) // reduced is RDD[((String, String), Int)]
val kv2 = reduced.map(e => e._1._1 -> (e._1._2 -> e._2)) // kv2 is RDD[(String, (String, Int))]
val grouped = kv2.groupByKey() // grouped is RDD[(String, Iterable[(String, Int)])]
grouped.foreach(println)
The syntax is below:
reduceByKey(func: Function2[V, V, V]): JavaPairRDD[K, V],
which says for the same key in an RDD it takes the values (which will be definitely of same type) performs the operation provided as part of function and returns the value of same type as of parent RDD.
Say I have been given a list of futures with each one linked to an key such as:
val seq: Seq[(Key, Future[Value])]
And my goal is to produce a list of key value tuples once all futures have completed:
val complete: Seq[(Key, Value)]
I am wondering if this can be achieved using a sequence call. For example I know I can do the following:
val complete = Future.sequence(seq.map(_._2).onComplete {
case Success(s) => s
case Failure(NonFatal(e)) => Seq()
}
But this will only returns me a sequence of Value objects and I lose the pairing information between Key and Value. The problem being that Future.sequence expects a sequence of Futures.
How could I augment this to maintain the key/value pairing in my complete sequence?
Thanks
Des
How about transforming your Seq[(Key, Future[Value])] to Seq[Future[(Key, Value)]] first.
val seq: Seq[(Key, Future[Value])] = // however your implementation is
val futurePair: Seq[Future[(Key, Value)]] = for {
(key, value) <- seq
} yield value.map(v => (key, v))
Now you can use sequence to get Future[Seq[(Key, Value)]].
val complete: Future[Seq[(String, Int)]] = Future.sequence(futurePair)
Just a different expression of the other answer, using unzip and zip.
scala> val vs = Seq(("one",Future(1)),("two",Future(2)))
vs: Seq[(String, scala.concurrent.Future[Int])] = List((one,scala.concurrent.impl.Promise$DefaultPromise#4e38d975), (two,scala.concurrent.impl.Promise$DefaultPromise#35f8a9d3))
scala> val (ks, fs) = vs.unzip
ks: Seq[String] = List(one, two)
fs: Seq[scala.concurrent.Future[Int]] = List(scala.concurrent.impl.Promise$DefaultPromise#4e38d975, scala.concurrent.impl.Promise$DefaultPromise#35f8a9d3)
scala> val done = (Future sequence fs) map (ks zip _)
done: scala.concurrent.Future[Seq[(String, Int)]] = scala.concurrent.impl.Promise$DefaultPromise#56913163
scala> done.value
res0: Option[scala.util.Try[Seq[(String, Int)]]] = Some(Success(List((one,1), (two,2))))
or maybe save on zippage:
scala> val done = (Future sequence fs) map ((ks, _).zipped)
done: scala.concurrent.Future[scala.runtime.Tuple2Zipped[String,Seq[String],Int,Seq[Int]]] = scala.concurrent.impl.Promise$DefaultPromise#766a52f5
scala> done.value.get.get.toList
res1: List[(String, Int)] = List((one,1), (two,2))