My requirement is that :
arr1 : Array[(String, String)] = Array((bangalore,Kanata), (kannur,Kerala))
arr2 : Array[(String, String)] = Array((001,anup), (002,sithu))
should give me
Array((001,anup,bangalore,Krnata), (002,sithu,kannur,Kerala))
I tried this :
val arr3 = arr2.map(field=>(field,arr1))
but it didn't work
#nicodp's answer addressed your question very nicely. zip and then map will give you the resultant array.
Recall that if one list is larger than the other, its remaining elements are ignored.
My attempt tries to address this:
Consider:
val arr1 = Array(("bangalore","Kanata"), ("kannur","Kerala"))
val arr2 = Array(("001","anup", "ramakrishan"), ("002","sithu", "bhattacharya"))
zip and mapping on tuples will give the result as:
arr1.zip(arr2).map(field => (field._1._1, field._1._2, field._2._1, field._2._2))
Array[(String, String, String, String)] = Array((bangalore,Kanata,001,anup), (kannur,Kerala,002,sithu))
// This ignores the last field of arr2
While mapping, you can convert the tuple in iterator and get a list from it. This will enable you to not keep a track of Tuple2 or Tuple3
arr1.zip(arr2).map{ case(k,v) => List(k.productIterator.toList, v.productIterator.toList).flatten }
// Array[List[Any]] = Array(List(bangalore, Kanata, 001, anup, ramakrishan), List(kannur, Kerala, 002, sithu, bhattacharya))
You can do a zip followed by a map:
scala> val arr1 = Array((1,2),(3,4))
arr1: Array[(Int, Int)] = Array((1,2), (3,4))
scala> val arr2 = Array((5,6),(7,8))
arr2: Array[(Int, Int)] = Array((5,6), (7,8))
scala> arr1.zip(arr2).map(field => (field._1._1, field._1._2, field._2._1, field._2._2))
res1: Array[(Int, Int, Int, Int)] = Array((1,2,5,6), (3,4,7,8))
The map acts as a flatten for tuples, that is, takes things of type ((A, B), (C, D)) and maps them to (A, B, C, D).
What zip does is... meh, let's see its type:
def zip[B](that: GenIterable[B]): List[(A, B)]
So, from there, we can argue that it takes an iterable collection (which can be another list) and returns a list which is the combination of the corresponding elements of both this: List[A] and that: List[B] lists. Recall that if one list is larger than the other, its remaining elements are ignored. You can dig more about list functions in the documentation.
I agree that the cleanes solution is using the zip method from collections
val arr1 = Array(("bangalore","Kanata"), ("kannur","Kerala"))
val arr2 = Array(("001","anup"), ("002","sithu"))
arr1.zip(arr2).foldLeft(List.empty[Any]) {
case (acc, (a, b)) => acc ::: List(a.productIterator.toList ++ b.productIterator.toList)
}
Related
My goal is to to map every word in a text (Index, line) to a list containing the indices of every line the word occurs in. I managed to write a function that returns a list of all words assigned to a index.
The following function should do the rest (map a list of indices to every word):
def mapIndicesToWords(l:List[(Int,String)]):Map[String,List[Int]] = ???
If I do this:
l.groupBy(x => x._2)
it returns a Map[String, List[(Int,String)]. Now I just want to change the value to type List[Int].
I thought of using .mapValues(...) and fold the list somehow, but I'm new to scala and don't know the correct approach for this.
So how do I convert the list?
Also you can use foldLeft, you need just specify accumulator (in your case Map[String, List[Int]]), which will be returned as a result, and write some logic inside. Here is my implementation.
def mapIndicesToWords(l:List[(Int,String)]): Map[String,List[Int]] =
l.foldLeft(Map[String, List[Int]]())((map, entry) =>
map.get(entry._2) match {
case Some(list) => map + (entry._2 -> (entry._1 :: list))
case None => map + (entry._2 -> List(entry._1))
}
)
But with foldLeft, elements of list will be in reversed order, so you can use foldRight. Just change foldLeft to foldRight and swap input parameters, (map, entry) to (entry, map).
And be careful, foldRight works 2 times slower. It is implemented using method reverse list and foldLeft.
scala> val myMap: Map[String,List[(Int, String)]] = Map("a" -> List((1,"line1"), (2, "line")))
myMap: Map[String,List[(Int, String)]] = Map(a -> List((1,line1), (2,line)))
scala> myMap.mapValues(lst => lst.map(pair => pair._1))
res0: scala.collection.immutable.Map[String,List[Int]] = Map(a -> List(1, 2))
Say I have a Iterable[(Int, String)]. How do I get an array of just the "values"? That is, how do I convert from Iterable[(Int, String)] => Array[String]? The "keys" or "values" do not have to be unique, and that's why I put them in quotation marks.
iterable.map(_._2).toArray
_._2 : take out the second element of the tuple represented by input variable( _ ) whose name I don't care.
Simply:
val iterable: Iterable[(Int, String)] = Iterable((1, "a"), (2, "b"))
val values = iterable.toArray.map(_._2)
Simply map the iterable and extract the second element(tuple._2),
scala> val iterable: Iterable[(Int, String)] = Iterable((100, "Bring me the horizon"), (200, "Porcupine Tree"))
iterable: Iterable[(Int, String)] = List((100,Bring me the horizon), (200,Porcupine Tree))
scala> iterable.map(tuple => tuple._2).toArray
res3: Array[String] = Array(Bring me the horizon, Porcupine Tree)
In addition to the already suggested map you might want to build the array as you map from tuple to string instead of converting at some point as it might save an iteration.
import scala.collection
val values: Array[String] = iterable.map(_._2)(collection.breakOut)
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))
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))
I am new to Scala and I am trying to figure out some scala syntax.
So I have a list of strings.
wordList: List[String] = List("this", "is", "a", "test")
I have a function that returns a list of pairs that contains consonants and vowels counts per word:
def countFunction(words: List[String]): List[(String, Int)]
So, for example:
countFunction(List("test")) => List(('Consonants', 3), ('Vowels', 1))
I now want to take a list of words and group them by count signatures:
def mapFunction(words: List[String]): Map[List[(String, Int)], List[String]]
//using wordList from above
mapFunction(wordList) => List(('Consonants', 3), ('Vowels', 1)) -> Seq("this", "test")
List(('Consonants', 1), ('Vowels', 1)) -> Seq("is")
List(('Consonants', 0), ('Vowels', 1)) -> Seq("a")
I'm thinking I need to use GroupBy to do this:
def mapFunction(words: List[String]): Map[List[(String, Int)], List[String]] = {
words.groupBy(F: (A) => K)
}
I've read the scala api for Map.GroupBy and see that F represents discriminator function and K is the type of keys you want returned. So I tried this:
words.groupBy(countFunction => List[(String, Int)]
However, scala doesn't like this syntax. I tried looking up some examples for groupBy and nothing seems to help me with my use case. Any ideas?
Based on your description, your count function should take a word instead of a list of words. I would have defined it like this:
def countFunction(words: String): List[(String, Int)]
If you do that you should be able to call words.groupBy(countFunction), which is the same as:
words.groupBy(word => countFunction(word))
If you cannot change the signature of countFunction, then you should be able to call group by like this:
words.groupBy(word => countFunction(List(word)))
You shouldn't put the return type of the function in the call. The compiler can figure this out itself. You should just call it like this:
words.groupBy(countFunction)
If that doesn't work, please post your countFunction implementation.
Update:
I tested it in the REPL and this works (note that my countFunction has a slightly different signature from yours):
scala> def isVowel(c: Char) = "aeiou".contains(c)
isVowel: (c: Char)Boolean
scala> def isConsonant(c: Char) = ! isVowel(c)
isConsonant: (c: Char)Boolean
scala> def countFunction(s: String) = (('Consonants, s count isConsonant), ('Vowels, s count isVowel))
countFunction: (s: String)((Symbol, Int), (Symbol, Int))
scala> List("this", "is", "a", "test").groupBy(countFunction)
res1: scala.collection.immutable.Map[((Symbol, Int), (Symbol, Int)),List[java.lang.String]] = Map((('Consonants,0),('Vowels,1)) -> List(a), (('Consonants,1),('Vowels,1)) -> List(is), (('Consonants,3),('Vowels,1)) -> List(this, test))
You can include the type of the function passed to groupBy, but like I said you don't need it. If you want to pass it in you do it like this:
words.groupBy(countFunction: String => ((Symbol, Int), (Symbol, Int)))