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

Scala 集合确实有 count: list.count(_ == 2)

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))

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

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

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

结果是:

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

值得注意的是，为这种情况特意设计的默认值为0的 map 表现出了最差的性能(并不像 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]) {
}

生产

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

令人好奇的是，最简洁的 groupBy甚至比可变的地图更快！

简短的回答:

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

长话短说:

使用 斯卡拉兹，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)

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))

投降

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

另一个答案的一个比较简洁的版本是:

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


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

给出一个 Map，对原始序列中的每个项目进行计数:

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:

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

还有一个选择:

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)

If you want to use it like list.count(2) you have to implement it using an 隐式课程.

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

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

给予

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

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

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

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

I did not get the size of the list using length but rather size as one the answer above suggested it because of the issue reported 给你.

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

使用 猫

import cats.implicits._


"Alphabet".toLowerCase().map(c => Map(c -> 1)).toList.combineAll
"Alphabet".toLowerCase().map(c => Map(c -> 1)).toList.foldMap(identity)

从 Scala 2.13开始，GroupMapReduce方法通过一次遍历列表来完成:

// 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

这个:

• group列出元素(小组MapReduce 的组部分)

• map的每个分组值出现到1(映射 group地图Reduce 的一部分)

• 通过求和(groupMap减速的 reduce 部分) ，在一组值(_ + _)中求出 reduce的值。

这是一个 一次过的版本，可以翻译成:

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

试试这个，应该有用。

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

它将返回3

这里有一个非常简单的方法来做到这一点。

val data = List("it", "was", "the", "best", "of", "times", "it", "was",
"the", "worst", "of", "times")
data.foldLeft(Map[String,Int]().withDefaultValue(0)){
case (acc, letter) =>
acc + (letter -> (1 + acc(letter)))
}
// => Map(worst -> 1, best -> 1, it -> 2, was -> 2, times -> 2, of -> 2, the -> 2)

val words = Array("Mary", "had", "a", "little", "lamb", "its"
, "fleece", "was", "white", "as", "snow", "and", "everywhere"
, "that", "Mary", "went", "the", "lamb", "was", "sure", "to", "go")




words.groupBy(_.length)