I have the following mutable Hashmap in Scala:
HashMap((b,3), (c,4), (a,8), (a,2))
and need to be converted to the following:
HashMap((b,3), (c,4), (a,10))
I need something like reduceByKey function logic.
I added the code here
def main(args: Array[String]) = {
val m = new mutable.HashMap[String,Tuple2[String,Int]]()
println("Hello, world")
m.+=(("xx",("a",2)))
m.+=(("uu",("b",3)))
m.+=(("zz",("a",8)))
m.+=(("yy",("c",4)))
println(m.values)
}
For pre 2.13 Scala versions you can try using groupBy with map:
m.values
.groupBy(_._1)
.mapValues(_.map(_._2).sum)
It sounds like what you have is not a hashmap but m.values of type Iterable[Tuple2[String, Int]], which is more manageable. In that case, as hinted at in the comments, groupMapReduce does it all in one function. This function groups "matching" elements together, applies a transformation to each element, and then reduces the groups using a binary operation.
m.values.groupMapReduce(_._1)(_._2)(_ + _)
This says "Group the values by the first element of their tuple, then keep the second element (i.e. the number), and then add all of the numbers in each group". This produces a map from the first element of the tuple to the sum.
Map(a -> 10, b -> 3, c -> 4)
Note that this is a Map, not necessarily a HashMap. If you want a HashMap (i.e. for mutability), you'll need to convert it yourself.
Related
I have two RDDs that wrap the following arrays:
Array((3,Ken), (5,Jonny), (4,Adam), (3,Ben), (6,Rhonda), (5,Johny))
Array((4,Rudy), (7,Micheal), (5,Peter), (5,Shawn), (5,Aaron), (7,Gilbert))
I need to design a code in such a way that if I provide input as 3 I need to return
Array((3,Ken), (3,Ben))
If input is 6, output should be
Array((6,Rhonda))
I tried something like this:
val list3 = list1.union(list2)
list3.reduceByKey(_+_).collect
list3.reduceByKey(6).collect
None of these worked, can anyone help me out with a solution for this problem?
Given the following that you would have to define yourself
// Provide you SparkContext and inputs here
val sc: SparkContext = ???
val array1: Array[(Int, String)] = ???
val array2: Array[(Int, String)] = ???
val n: Int = ???
val rdd1 = sc.parallelize(array1)
val rdd2 = sc.parallelize(array2)
You can use the union and filter to reach your goal
rdd1.union(rdd2).filter(_._1 == n)
Since filtering by key is something that you would probably want to do in several occasions, it makes sense to encapsulate this functionality in its own function.
It would also be interesting if we could make sure that this function could work on any type of keys, not only Ints.
You can express this in the old RDD API as follows:
def filterByKey[K, V](rdd: RDD[(K, V)], k: K): RDD[(K, V)] =
rdd.filter(_._1 == k)
You can use it as follows:
val rdd = rdd1.union(rdd2)
val filtered = filterByKey(rdd, n)
Let's look at this method a little bit more in detail.
This method allows to filterByKey and RDD which contains a generic pair, where the type of the first item is K and the type of the second type is V (from key and value). It also accepts a key of type K that will be used to filter the RDD.
You then use the filter function, that takes a predicate (a function that goes from some type - in this case K - to a Boolean) and makes sure that the resulting RDD only contains items that respect this predicate.
We could also have written the body of the function as:
rdd.filter(pair => pair._1 == k)
or
rdd.filter { case (key, value) => key == k }
but we took advantage of the _ wildcard to express the fact that we want to act on the first (and only) parameter of this anonymous function.
To use it, you first parallelize your RDDs, call union on them and then invoke the filterByKey function with the number you want to filter by (as shown in the example).
I am a newbie in scala and I need to sort a very large list with 40000 integers.
The operation is performed many times. So performance is very important.
What is the best method for sorting?
You can sort the list with List.sortWith() by providing a relevant function literal. For example, the following code prints all elements of sorted list which contains all elements of the initial list in alphabetical order of the first character lowercased:
val initial = List("doodle", "Cons", "bible", "Army")
val sorted = initial.sortWith((s: String, t: String)
=> s.charAt(0).toLower < t.charAt(0).toLower)
println(sorted)
Much shorter version will be the following with Scala's type inference:
val initial = List("doodle", "Cons", "bible", "Army")
val sorted = initial.sortWith((s, t) => s.charAt(0).toLower < t.charAt(0).toLower)
println(sorted)
For integers there is List.sorted, just use this:
val list = List(4, 3, 2, 1)
val sortedList = list.sorted
println(sortedList)
just check the docs
List has several methods for sorting. myList.sorted works for types with already defined order (like Int or String and others). myList.sortWith and myList.sortBy receive a function that helps defining the order
Also, first link on google for scala List sort: http://alvinalexander.com/scala/how-sort-scala-sequences-seq-list-array-buffer-vector-ordering-ordered
you can use List(1 to 400000).sorted
I have an emptyListBuffer[ListBuffer[(String, Int)]]() initialized like so, and given a number n, I want to fill it with n ListBuffer[(String, Int)].
For example, if n=2 then I can initialize two ListBuffer[(String, Int)] within ListBuffer[ListBuffer[(String, Int)]]() if that makes any sense. I was trying to loop n times and use the insertAll function to insert an empty list but I didn't work.
use fill
fill is a standard Scala library function in order to fill a data structure with predefined elements. Its quite handy and save lot of typing.
ListBuffer.fill(100)(ListBuffer("Scala" -> 1))
Scala REPL
scala> import scala.collection.mutable._
import scala.collection.mutable._
scala> ListBuffer.fill(100)(ListBuffer("Scala" -> 1))
res4: scala.collection.mutable.ListBuffer[scala.collection.mutable.ListBuffer[(String, Int)]] = ListBuffer(ListBuffer((Scala,1)), ListBuffer((Scala,1)), ListBuffer((Scala,1)), ListBuffer((Scala,1)), ListBuffer((Scala,1)) ...
fill implementation in Standard library
def fill[A](n: Int)(elem: => A): CC[A] = {
val b = newBuilder[A]
b.sizeHint(n)
var i = 0
while (i < n) {
b += elem
i += 1
}
b.result()
}
The above implementation is for one dimensional data structure.
General suggestions
Looks like you are using Scala like the Java way. This is not good. Embrace functional way for doing things for obvious benefits.
Use immutable collections like List, Vector instead of mutable collections. Do not use mutable collections until and unless you have string reason for it.
Same thing can be done using immutable List
List.fill(100)(List("scala" -> 1))
scala -> 1 is same as ("scala", 1)
I was trying to find line with maximum words, and i wrote the following lines, to run on spark-shell:
import java.lang.Math
val counts = textFile.map(line => line.split(" ").size).reduce((a, b) => Math.max(a, b))
But since, map is one to one , and flatMap is one to either zero or anything. So i tried replacing map with flatMap, in above code. But its giving error as:
<console>:24: error: type mismatch;
found : Int
required: TraversableOnce[?]
val counts = F1.flatMap(s => s.split(" ").size).reduce((a,b)=> Math.max(a,b))
If anybody could make me understand the reason, it will really be helpful.
flatMap must return an Iterable which is clearly not what you want. You do want a map because you want to map a line to the number of words, so you want a one-to-one function that takes a line and maps it to the number of words (though you could create a collection with one element, being the size of course...).
FlatMap is meant to associate a collection to an input, for instance if you wanted to map a line to all its words you would do:
val words = textFile.flatMap(x => x.split(" "))
and that would return an RDD[String] containing all the words.
In the end, map transforms an RDD of size N into another RDD of size N (e.g. your lines to their length) whereas flatMap transforms an RDD of size N into an RDD of size P (actually an RDD of size N into an RDD of size N made of collections, all these collections are then flattened to produce the RDD of size P).
P.S.: one last word that has nothing to do with your problem, it is more efficient to do (for a string s)
val nbWords = s.split(" ").length
than call .size(). Indeed, the split method returns an array of String and arrays do not have a size method. So when you call .size() you have an implicit conversion from Array[String] to SeqLike[String] which creates new objects. But Array[T] do have a length field so there's no conversion calling length. (It's a detail but I think it's good habit though).
Any use of map can be replaced by flatMap, but the function argument has to be changed to return a single-element List: textFile.flatMap(line => List(line.split(" ").size)). This isn't a good idea: it just makes your code less understandable and less efficient.
After reading Tired of Null Pointer Exceptions? Consider Using Java SE 8's Optional!'s part about why use flatMap() rather than Map(), I have realized the truly reason why flatMap() can not replace map() is that map() is not a special case of flatMap().
It's true that flatMap() means one-to-many, but that's not the only thing flatMap() does. It can also strip outer Stream() if put it simply.
See the definations of map and flatMap:
Stream<R> map(Function<? super T, ? extends R> mapper)
Stream<R> flatMap(Function<? super T, ? extends Stream<? extends R>> mapper)
the only difference is the type of returned value in inner function. What map() returned is "Stream<'what inner function returned'>", while what flatMap() returned is just "what inner function returned".
So you can say that flatMap() can kick outer Stream() away, but map() can't. This is the most difference in my opinion, and also why map() is not just a special case of flatMap().
ps:
If you really want to make one-to-one with flatMap, then you should change it into one-to-List(one). That means you should add an outer Stream() manually which will be stripped by flatMap() later. After that you'll get the same effect as using map().(Certainly, it's clumsy. So don't do like that.)
Here are examples for Java8, but the same as Scala:
use map():
list.stream().map(line -> line.split(" ").length)
deprecated use flatMap():
list.stream().flatMap(line -> Arrays.asList(line.split(" ").length).stream())
I have a for-comprehension with a generator from a Set[MyType]
This MyType has a lazy val variable called factsPair which returns a pair of sets:
(Set[MyFact], Set[MyFact]).
I wish to loop through all of them and unify the facts into one flattened pair (Set[MyFact], Set[MyFact]) as follows, however I am getting No implicit view available ... and not enough arguments for flatten: implicit (asTraversable ... errors. (I am a bit new to Scala so still trying to get used to the errors).
lazy val allFacts =
(for {
mytype <- mytypeList
} yield mytype.factsPair).flatten
What do I need to specify to flatten for this to work?
Scala flatten works on same types. You have a Seq[(Set[MyFact], Set[MyFact])], which can't be flattened.
I would recommend learning the foldLeft function, because it's very general and quite easy to use as soon as you get the hang of it:
lazy val allFacts = myTypeList.foldLeft((Set[MyFact](), Set[MyFact]())) {
case (accumulator, next) =>
val pairs1 = accumulator._1 ++ next.factsPair._1
val pairs2 = accumulator._2 ++ next.factsPair._2
(pairs1, pairs2)
}
The first parameter takes the initial element it will append the other elements to. We start with an empty Tuple[Set[MyFact], Set[MyFact]] initialized like this: (Set[MyFact](), Set[MyFact]()).
Next we have to specify the function that takes the accumulator and appends the next element to it and returns with the new accumulator that has the next element in it. Because of all the tuples, it doesn't look nice, but works.
You won't be able to use flatten for this, because flatten on a collection returns a collection, and a tuple is not a collection.
You can, of course, just split, flatten, and join again:
val pairs = for {
mytype <- mytypeList
} yield mytype.factsPair
val (first, second) = pairs.unzip
val allFacts = (first.flatten, second.flatten)
A tuple isn't traverable, so you can't flatten over it. You need to return something that can be iterated over, like a List, for example:
List((1,2), (3,4)).flatten // bad
List(List(1,2), List(3,4)).flatten // good
I'd like to offer a more algebraic view. What you have here can be nicely solved using monoids. For each monoid there is a zero element and an operation to combine two elements into one.
In this case, sets for a monoid: the zero element is an empty set and the operation is a union. And if we have two monoids, their Cartesian product is also a monoid, where the operations are defined pairwise (see examples on Wikipedia).
Scalaz defines monoids for sets as well as tuples, so we don't need to do anything there. We'll just need a helper function that combines multiple monoid elements into one, which is implemented easily using folding:
def msum[A](ps: Iterable[A])(implicit m: Monoid[A]): A =
ps.foldLeft(m.zero)(m.append(_, _))
(perhaps there already is such a function in Scala, I didn't find it). Using msum we can easily define
def pairs(ps: Iterable[MyType]): (Set[MyFact], Set[MyFact]) =
msum(ps.map(_.factsPair))
using Scalaz's implicit monoids for tuples and sets.