斯卡拉我如何计算列表中出现的次数

val list = List(1,2,4,2,4,7,3,2,4) 

我想实现它像这样： list.count(2) （返回3）。

Answer 1

其他答案之一的更清晰的版本是：

 val s = Seq("apple", "oranges", "apple", "banana", "apple", "oranges", "oranges") s.groupBy(identity).mapValues(_.size) 

为原始序列中的每个项目分配一个计数：

 Map(banana -> 1, oranges -> 3, apple -> 3) 

这个问题问如何find一个特定项目的数量。 采用这种方法，解决scheme将需要将期望的元素映射到其计数值，如下所示：

 s.groupBy(identity).mapValues(_.size)("apple") 

Answer 2

斯卡拉集合确实有count ： list.count(_ == 2)

Answer 3

我遇到了和Sharath Prabhal一样的问题，而我得到了另一个（对我来说更清楚）的解决scheme：

 val s = Seq("apple", "oranges", "apple", "banana", "apple", "oranges", "oranges") s.groupBy(l => l).map(t => (t._1, t._2.length)) 

结果是：

 Map(banana -> 1, oranges -> 3, apple -> 3) 

Answer 4

 val list = List(1, 2, 4, 2, 4, 7, 3, 2, 4) // Using the provided count method this would yield the occurrences of each value in the list: l map(x => l.count(_ == x)) List[Int] = List(1, 3, 3, 3, 3, 1, 1, 3, 3) // This will yield a list of pairs where the first number is the number from the original list and the second number represents how often the first number occurs in the list: l map(x => (x, l.count(_ == x))) // outputs => List[(Int, Int)] = List((1,1), (2,3), (4,3), (2,3), (4,3), (7,1), (3,1), (2,3), (4,3)) 

Answer 5

 list.groupBy(i=>i).mapValues(_.size) 

给

 Map[Int,Int] = Map(1 -> 1, 2 -> 3, 7 -> 1, 3 -> 1, 4 -> 3) 

Answer 6

如果你想使用它像list.count(2)你必须使用一个隐式类实现它。

 implicit class Count[T](list: List[T]) { def count(n: T): Int = list.count(_ == n) } List(1,2,4,2,4,7,3,2,4).count(2) // returns 3 List(1,2,4,2,4,7,3,2,4).count(5) // returns 0 

Answer 7

有趣的是，值得注意的是，默认情况下针对这种情况devise的映射performance出最差的性能（而不像groupBy那样简洁）

  type Word = String type Sentence = Seq[Word] type Occurrences = scala.collection.Map[Char, Int] def woGrouped(w: Word): Occurrences = { w.groupBy(c => c).map({case (c, list) => (c -> list.length)}) } //> woGrouped: (w: forcomp.threadBug.Word)forcomp.threadBug.Occurrences def woGetElse0Map(w: Word): Occurrences = { val map = Map[Char, Int]() w.foldLeft(map)((m, c) => m + (c -> (m.getOrElse(c, 0) + 1)) ) } //> woGetElse0Map: (w: forcomp.threadBug.Word)forcomp.threadBug.Occurrences def woDeflt0Map(w: Word): Occurrences = { val map = Map[Char, Int]().withDefaultValue(0) w.foldLeft(map)((m, c) => m + (c -> (m(c) + 1)) ) } //> woDeflt0Map: (w: forcomp.threadBug.Word)forcomp.threadBug.Occurrences def dfltHashMap(w: Word): Occurrences = { val map = scala.collection.immutable.HashMap[Char, Int]().withDefaultValue(0) w.foldLeft(map)((m, c) => m + (c -> (m(c) + 1)) ) } //> dfltHashMap: (w: forcomp.threadBug.Word)forcomp.threadBug.Occurrences def mmDef(w: Word): Occurrences = { val map = scala.collection.mutable.Map[Char, Int]().withDefaultValue(0) w.foldLeft(map)((m, c) => m += (c -> (m(c) + 1)) ) } //> mmDef: (w: forcomp.threadBug.Word)forcomp.threadBug.Occurrences val functions = List("grp" -> woGrouped _, "mtbl" -> mmDef _, "else" -> woGetElse0Map _ , "dfl0" -> woDeflt0Map _, "hash" -> dfltHashMap _ ) //> functions : List[(String, String => scala.collection.Map[Char,Int])] = Lis //| t((grp,<function1>), (mtbl,<function1>), (else,<function1>), (dfl0,<functio //| n1>), (hash,<function1>)) val len = 100 * 1000 //> len : Int = 100000 def test(len: Int) { val data: String = scala.util.Random.alphanumeric.take(len).toList.mkString val firstResult = functions.head._2(data) def run(f: Word => Occurrences): Int = { val time1 = System.currentTimeMillis() val result= f(data) val time2 = (System.currentTimeMillis() - time1) assert(result.toSet == firstResult.toSet) time2.toInt } def log(results: Seq[Int]) = { ((functions zip results) map {case ((title, _), r) => title + " " + r} mkString " , ") } var groupResults = List.fill(functions.length)(1) val integrals = for (i <- (1 to 10)) yield { val results = functions map (f => (1 to 33).foldLeft(0) ((acc,_) => run(f._2))) println (log (results)) groupResults = (results zip groupResults) map {case (r, gr) => r + gr} log(groupResults).toUpperCase } integrals foreach println } //> test: (len: Int)Unit test(len) test(len * 2) // GRP 14 , mtbl 11 , else 31 , dfl0 36 , hash 34 // GRP 91 , MTBL 111 println("Done") def main(args: Array[String]) { } 

产生

 grp 5 , mtbl 5 , else 13 , dfl0 17 , hash 17 grp 3 , mtbl 6 , else 14 , dfl0 16 , hash 16 grp 3 , mtbl 6 , else 13 , dfl0 17 , hash 15 grp 4 , mtbl 5 , else 13 , dfl0 15 , hash 16 grp 23 , mtbl 6 , else 14 , dfl0 15 , hash 16 grp 5 , mtbl 5 , else 13 , dfl0 16 , hash 17 grp 4 , mtbl 6 , else 13 , dfl0 16 , hash 16 grp 4 , mtbl 6 , else 13 , dfl0 17 , hash 15 grp 3 , mtbl 5 , else 14 , dfl0 16 , hash 16 grp 3 , mtbl 6 , else 14 , dfl0 16 , hash 16 GRP 5 , MTBL 5 , ELSE 13 , DFL0 17 , HASH 17 GRP 8 , MTBL 11 , ELSE 27 , DFL0 33 , HASH 33 GRP 11 , MTBL 17 , ELSE 40 , DFL0 50 , HASH 48 GRP 15 , MTBL 22 , ELSE 53 , DFL0 65 , HASH 64 GRP 38 , MTBL 28 , ELSE 67 , DFL0 80 , HASH 80 GRP 43 , MTBL 33 , ELSE 80 , DFL0 96 , HASH 97 GRP 47 , MTBL 39 , ELSE 93 , DFL0 112 , HASH 113 GRP 51 , MTBL 45 , ELSE 106 , DFL0 129 , HASH 128 GRP 54 , MTBL 50 , ELSE 120 , DFL0 145 , HASH 144 GRP 57 , MTBL 56 , ELSE 134 , DFL0 161 , HASH 160 grp 7 , mtbl 11 , else 28 , dfl0 31 , hash 31 grp 7 , mtbl 10 , else 28 , dfl0 32 , hash 31 grp 7 , mtbl 11 , else 28 , dfl0 31 , hash 32 grp 7 , mtbl 11 , else 28 , dfl0 31 , hash 33 grp 7 , mtbl 11 , else 28 , dfl0 32 , hash 31 grp 8 , mtbl 11 , else 28 , dfl0 31 , hash 33 grp 8 , mtbl 11 , else 29 , dfl0 38 , hash 35 grp 7 , mtbl 11 , else 28 , dfl0 32 , hash 33 grp 8 , mtbl 11 , else 32 , dfl0 35 , hash 41 grp 7 , mtbl 13 , else 28 , dfl0 33 , hash 35 GRP 7 , MTBL 11 , ELSE 28 , DFL0 31 , HASH 31 GRP 14 , MTBL 21 , ELSE 56 , DFL0 63 , HASH 62 GRP 21 , MTBL 32 , ELSE 84 , DFL0 94 , HASH 94 GRP 28 , MTBL 43 , ELSE 112 , DFL0 125 , HASH 127 GRP 35 , MTBL 54 , ELSE 140 , DFL0 157 , HASH 158 GRP 43 , MTBL 65 , ELSE 168 , DFL0 188 , HASH 191 GRP 51 , MTBL 76 , ELSE 197 , DFL0 226 , HASH 226 GRP 58 , MTBL 87 , ELSE 225 , DFL0 258 , HASH 259 GRP 66 , MTBL 98 , ELSE 257 , DFL0 293 , HASH 300 GRP 73 , MTBL 111 , ELSE 285 , DFL0 326 , HASH 335 Done 

奇怪的是，最简洁的groupBy比甚至是可变的地图还要快！

Answer 8

简短的回答：

 import scalaz._, Scalaz._ xs.foldMap(x => Map(x -> 1)) 

很长的回答：

使用Scalaz ，例如

 import scalaz._, Scalaz._ val xs = List('a, 'b, 'c, 'c, 'a, 'a, 'b, 'd) 

那么所有这些（按照简化到简化的顺序）

 xs.map(x => Map(x -> 1)).foldMap(identity) xs.map(x => Map(x -> 1)).foldMap() xs.map(x => Map(x -> 1)).suml xs.map(_ -> 1).foldMap(Map(_)) xs.foldMap(x => Map(x -> 1)) 

产量

 Map('b -> 2, 'a -> 3, 'c -> 2, 'd -> 1) 

Answer 9

我遇到了同样的问题，但想要一次统计多个项目..

 val s = Seq("apple", "oranges", "apple", "banana", "apple", "oranges", "oranges") s.foldLeft(Map.empty[String, Int]) { (m, x) => m + ((x, m.getOrElse(x, 0) + 1)) } res1: scala.collection.immutable.Map[String,Int] = Map(apple -> 3, oranges -> 3, banana -> 1) 

https://gist.github.com/sharathprabhal/6890475

Answer 10

这是另一个select：

 scala> val list = List(1,2,4,2,4,7,3,2,4) list: List[Int] = List(1, 2, 4, 2, 4, 7, 3, 2, 4) scala> list.groupBy(x => x) map { case (k,v) => k-> v.length } res74: scala.collection.immutable.Map[Int,Int] = Map(1 -> 1, 2 -> 3, 7 -> 1, 3 -> 1, 4 -> 3) 

Answer 11

 scala> val list = List(1,2,4,2,4,7,3,2,4) list: List[Int] = List(1, 2, 4, 2, 4, 7, 3, 2, 4) scala> println(list.filter(_ == 2).size) 3 

Answer 12

我没有得到length的列表的大小，而是size作为一个上面的答案build议，因为这里报道的问题。

 val s = List("apple", "oranges", "apple", "banana", "apple", "oranges", "oranges") list.groupBy(x=>x).map(t => (t._1, t._2.size))