Scala简单直方图

例如，对于给定的

数组[Double]

val a = Array.tabulate(100){ _ => Random.nextDouble * 10 }

使用

n

bin计算直方图的简单方法是什么

这个怎么样

object Hist {

    type Bins = Map[Double, List[Double]]
    // artificially increasing bucket length to overcome last-point issue 
    private val Epsilon = 0.000001

    def histogram(data: List[Double], binsCount: Int) = {
        require(data.length > binsCount)
        val sorted = data.sorted
        val min = sorted.head
        val max = sorted.last
        val binLength = (max - min) / binsCount + Epsilon

        val bins = Map.empty[Double, List[Double]].withDefaultValue(Nil)

        scatterToBins(sorted, min + binLength, binLength, bins)
    }

    @annotation.tailrec
    private def scatterToBins(xs: List[Double], upperBound: Double, binLength: Double, bins: Bins): Bins = xs match {
        case Nil         => bins
        case point::tail => 
            val bound = if (point < upperBound) upperBound else upperBound + binLength
            val currentBin = bins(bound)
            val newBin = point::currentBin
            scatterToBins(tail, bound, binLength, bins + (bound -> newBin))             
    }

    // now let's test this out
    val data = Array.tabulate(100){ _ => scala.util.Random.nextDouble * 10 }

    val result = histogram(data.toList, 5)

    val pointsPerBucket = result.values.map(xs => xs.length)
}

---

我用列表而不是数组欺骗了一些人，但我希望这对你没问题。

一种与@om-nom-nom的答案中非常相似的值准备方法，但是使用分区的直方图方法非常小

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 = (1 to nBins).map { x => min + binWidth * x }.toList

  def histo(bounds: List[Double], data: List[Double]): List[List[Double]] =
    bounds match {
      case h :: Nil => List(data)
      case h :: t   => val (l,r) = data.partition( _ < h) ; l :: histo(t,r)
    }

  val histogram = histo(bounds, data)
}

诸如此类

val tabulated = h.map {_.size}

这个怎么样：

val num_bins = 20
val mx = a.max.toDouble
val mn = a.min.toDouble
val hist = a
    .map(x=>(((x.toDouble-mn)/(mx-mn))*num_bins).floor.toInt)
    .groupBy(x=>x)
    .map(x=>x._1->x._2.size)
    .toSeq
    .sortBy(x=>x._1)
    .map(x=>x._2)

---

另一个答案，在我看来更简洁

def mkHistogram(n_bins: Int, lowerUpperBound: Option[(Double, Double)] = None)(xs: Seq[Double]) = {
  val (mn, mx) = lowerUpperBound getOrElse(xs.min, xs.max)
  val epsilon = 0.0001
  val binSize = (mx - mn) / n_bins * (1 + epsilon)
  val bins = (0 to n_bins).map(mn + _ * binSize).sliding(2).map(xs => (xs(0), xs(1)))
  def binContains(bin:(Double,Double),x: Double) = (x >= bin._1) && (x < bin._2)
  bins.map(bin => (bin, xs.count(binContains(bin,_))))
}


@ mkHistogram(5,Option(0,10))(Seq(1,1,1,1,2,2,2,3,4,5,6,7)).foreach(println)
((0.0,2.0002),7)
((2.0002,4.0004),2)
((4.0004,6.0006),2)
((6.0006,8.0008),1)
((8.0008,10.001),0)

def mkHistogram（n_bins:Int，loweruperbound:Option[（Double，Double）]=None）（xs:Seq[Double]）{
val（mn，mx）=下限上限getOrElse（xs.min，xs.max）
valε=0.0001
val binSize=（mx-mn）/n_bins*（1+ε）
val bins=（0到n个bins）.map（mn+*binSize）.slideing（2）.map（xs=>（xs（0），xs（1）））
def binContains（bin:（Double，Double），x:Double）=（x>=bin.\u 1）和&（x（bin，xs.count（binContains，bin，）））
}
@Mk直方图（5，选项（0,10））（序号（1,1,1,2,2,3,4,5,6,7））.foreach（println）
((0.0,2.0002),7)
((2.0002,4.0004),2)
((4.0004,6.0006),2)
((6.0006,8.0008),1)
((8.0008,10.001),0)

我有一个类似但略有不同的要求——根据用户定义的箱子/截止值制作直方图。比如，在OP的例子中，需要0-3、-4、-5、-6、-7,8+个箱子。我尝试了几种方法，但我的突破是认识到我需要根据每个值所插入的bin中的位置对数组进行分组：

val a = Array.tabulate(100){ _ => Random.nextDouble * 10 }
val bins=List(3,4,5,6,7,8,Int.MaxValue) //-- user-defined cutoff values (with max value at the top)
a.groupBy(i => bins.indexWhere(_>i)) //-- collection of lists fitting this criteria
  .map{case (i,items) => i -> items.length} //-- map for index to number of items in that index's list

在这种情况下，结果是：

Map(0 -> 26, 5 -> 7, 1 -> 5, 6 -> 24, 2 -> 12, 3 -> 15, 4 -> 11)

val a = Array.tabulate(100){ _ => Random.nextDouble * 10 }
val bins=List(3,4,5,6,7,8,Int.MaxValue) //-- user-defined cutoff values (with max value at the top)
a.groupBy(i => bins.indexWhere(_>i)) //-- collection of lists fitting this criteria
  .map{case (i,items) => i -> items.length} //-- map for index to number of items in that index's list

Map(0 -> 26, 5 -> 7, 1 -> 5, 6 -> 24, 2 -> 12, 3 -> 15, 4 -> 11)