Scala 如何计算列表中出现的次数

Scala how can I count the number of occurrences in a list

提问by Gatspy

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

I want to implement it like this: list.count(2)(returns 3).

我想像这样实现它：（list.count(2)返回 3）。

回答by ohruunuruus

A somewhat cleaner version of one of the other answers is:

其他答案之一的更简洁版本是：

val s = Seq("apple", "oranges", "apple", "banana", "apple", "oranges", "oranges")

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

giving a Mapwith a count for each item in the original sequence:

为Map原始序列中的每个项目提供一个计数：

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

The question asks how to find the count of a specific item. With this approach, the solution would require mapping the desired element to its count value as follows:

该问题询问如何查找特定项目的计数。使用这种方法，解决方案需要将所需元素映射到其计数值，如下所示：

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

回答by xiefei

scala collections do have count: list.count(_ == 2)

Scala 集合确实有count：list.count(_ == 2)

回答by KWA

I had the same problem as Sharath Prabhal, and I got another (to me clearer) solution :

我和 Sharath Prabhal 遇到了同样的问题，我得到了另一个（对我来说更清晰）的解决方案：

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

With as result :

结果是：

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

回答by noego

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

gives

给

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

Note that you can replace (i=>i)with built in identityfunction:

请注意，您可以(i=>i)使用内置identity函数替换：

list.groupBy(identity).mapValues(_.size)

回答by AndreasScheinert

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))

回答by Xavier Guihot

Starting Scala 2.13, the groupMapReducemethod does that in one pass through the list:

从一开始Scala 2.13，groupMapReduce方法就完成了对列表的一次传递：

// val seq = Seq("apple", "oranges", "apple", "banana", "apple", "oranges", "oranges")
seq.groupMapReduce(identity)(_ => 1)(_ + _)
// immutable.Map[String,Int] = Map(banana -> 1, oranges -> 3, apple -> 3)
seq.groupMapReduce(identity)(_ => 1)(_ + _)("apple")
// Int = 3

This:

这：

• groups list elements (group part of groupMapReduce)

• maps each grouped value occurrence to 1 (map part of groupMapReduce)

• reduces values within a group of values (_ + _) by summing them (reduce part of groupMapReduce).

• groups 列表元素（组MapReduce 的组部分）

• maps 每个分组值出现为 1（组MapReduce 的映射部分）

• reduce_ + _通过对一组值 ( ) 中的值进行求和（减少 groupMap Reduce 的一部分）。

This is a one-pass versionof what can be translated by:

这是可以通过以下方式翻译的内容的一次性版本：

seq.groupBy(identity).mapValues(_.map(_ => 1).reduce(_ + _))

回答by Sharath Prabhal

I ran into the same problem but wanted to count multiple items in one go..

我遇到了同样的问题，但想一次计算多个项目。

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

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

回答by Erik Kaplun

Short answer:

简答：

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

Long answer:

长答案：

Using Scalaz, given.

使用Scalaz，给定。

import scalaz._, Scalaz._

val xs = List('a, 'b, 'c, 'c, 'a, 'a, 'b, 'd)

then all of these (in the order of less simplified to more simplified)

然后所有这些（按照从简化到简化的顺序）

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))

yield

屈服

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

回答by LRLucena

If you want to use it like list.count(2)you have to implement it using an Implicit Class.

如果您想使用它，就像list.count(2)必须使用Implicit Class来实现它一样。

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

回答by Val

It is interesting to note that the map with default 0 value, intentionally designed for this case demonstrates the worst performance (and not as concise as groupBy)

有趣的是，默认为 0 值的地图，为这种情况而设计的，表现出最差的性能（而且不像 那样简洁groupBy）

    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]) {
    }

produces

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

It is curious that most concise groupByis faster than even mutable map!

奇怪的是，最简洁groupBy甚至比可变地图还要快！