scala-collections

问题 Does Scala provide a built-in class, utility, syntax, or other mechanism for converting (by wrapping) an Iterator with an Iterable? For example, I have an Iterator[Foo] and I need an Iterable[Foo], so currently I am: val foo1: Iterator[Foo] = .... val foo2: Iterable[Foo] = new Iterable[Foo] { def elements = foo1 } This seems ugly and unnecessary. What's a better way? 回答1: Iterator has a toIterable method in Scala 2.8.0, but not in 2.7.7 or earlier. It's not implicit, but you could define your

问题 This is a minor issue with one of my spark jobs which doesn't seem to cause any issues -- yet annoys me every time I see it and fail to come up with a better solution. Say I have a Scala collection like this: val myStuff = List(Try(2/2), Try(2/0)) I can partition this list into successes and failures with partition: val (successes, failures) = myStuff.partition(_.isSuccess) Which is nice. The implementation of partition only traverses the source collection once to build the two new

问题 Is there any methods available to remove every n th element from the Scala List ? I hope we can do this inside filter method and return another list by writing a logic. But is this efficient way to do it? 回答1: Simply: list.zipWithIndex .filter { case (_, i) => (i + 1) % n != 0 } .map { case (e, _) => e } 回答2: Simplest so far, I think def removeNth[A](myList: List[A], n: Int): List[A] = myList.zipWithIndex collect { case (x,i) if (i + 1) % n != 0 => x } collect is an oft-forgotten gem that

问题 I have a List val first = List("A","B","C","D") and I want to create a new list from it but change the last element only: val newLastVal = "E" val second = List("A","B","C","E") can't figure this one out! Thanks in advance 回答1: you can also use .init or .dropRight(1) to remove last element and then can add new item to list val second=first.init:+newLastVal //preferable OR val second=first.dropRight(1):+newLastVal 回答2: you can use .updated(postion,value) val second=first.updated(first.length-1

问题 Why does the immutable version of the ListMap store in ascending order, while mutable version stores in descending order? Here is a test that you can use if you got scalatest-1.6.1.jar and junit-4.9.jar @Test def StackoverflowQuestion() { val map = Map("A" -> 5, "B" -> 12, "C" -> 2, "D" -> 9, "E" -> 18) val sortedIMMUTABLEMap = collection.immutable.ListMap[String, Int](map.toList.sortBy[Int](_._2): _*) println("head : " + sortedIMMUTABLEMap.head._2) println("last : " + sortedIMMUTABLEMap.last

问题 I'm looking at using a DoubleLinkedList. It's remove() method says "Removes the current node from the double linked list." but there are no other references to current in the page. What is the current node, how do I set it, and surely this can't be the only way of removing an item? 回答1: A DoubleLinkedList is at the same time the list itself and a list node, similar to the :: for a regular List . You can navigate from one cell to the next or to the previous one with next and prev ,

问题 Assume you have a List(1,"1") it is typed List[Any], which is of course correct and expected. Now if I map the list like this scala> List(1, "1") map { | case x: Int => x | case y: String => y.toInt | } the resulting type is List[Int] which is expected as well. My question is if there is an equivalent to map for filter because the following example will result in a List[Any]. Is this possible? I assume this could be solved at compile time and possibly not runtime? scala> List(1, "1") filter {

问题 What is the most succinct Scala way to reverse a Map? The Map may contain non-unique values. EDIT: The reversal of Map[A, B] should give Map[B, Set[A]] (or a MultiMap, that would be even better). 回答1: If you can lose duplicate keys: scala> val map = Map(1->"one", 2->"two", -2->"two") map: scala.collection.immutable.Map[Int,java.lang.String] = Map((1,one), (2,two), (-2,two)) scala> map.map(_ swap) res0: scala.collection.immutable.Map[java.lang.String,Int] = Map((one,1), (two,-2)) If you don't

问题 Following Scala courses on Coursera, Martin Odersky showed an example code which is: 1 to 5 map ( i => i*i ) And he said the Range gets transformed to a Vector because they share the same interface ( IndexedSeq ) and the result could not be represented as a Range (it was more clear in its example since he generated a pair which is not representable as a Range ). I'm not sure to understand because I think he said previously that in a for expression the 1st generator will determine the kind of

问题 For a given Array[Double] , for instance val a = Array.tabulate(100){ _ => Random.nextDouble * 10 } what is a simple approach to calculate a histogram with n bins ? 回答1: A very similar preparation of values as in @om-nom-nom 's answer, yet the histogram method quite small by using partition , case class Distribution(nBins: Int, data: List[Double]) { require(data.length > nBins) val Epsilon = 0.000001 val (max,min) = (data.max,data.min) val binWidth = (max - min) / nBins + Epsilon val bounds =