Question

对于给定的Array[Double]，例如

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

用n箱计算直方图的简单方法是什么？

Answer 1

与@ om-nom-nom的答案非常相似的值准备，但直方图方法使用partition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)
}

然后

val data = Array.tabulate(100){ _ => scala.util.Random.nextDouble * 10 }
val h = Distribution(5, data.toList).histogram

等等

val tabulated = h.map {_.size}

Answer 2

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)

Answer 3

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)
}

产生以下输出：

scala> Hist.result
// res14: Hist.Bins = Map(4.043605797342332 -> List(4.031739029821568, 3.826704675600351, 3.7661438110766166, 3.680326808626887, 3.6788463836133767, 3.5442867825350266, 3.5156167603774904, 3.464310876575163, 3.3796397333178216, 3.33851670739545, 3.1702423754536504, 3.1681320879333708, 2.9520859637868204, 2.885027245987456, 2.8091011617711024, 2.745475619527371, 2.520275275070399, 2.3720116613386546, 2.2909255324112374, 2.229522549904405, 2.0693233045454895), 6.0237846547671845 -> List(5.957572654029027, 5.6887311125180675, 5.356707271645041, 5.3155138169898475, 5.285634121992783, 5.2823949256676865, 5.159891625116016, 5.152024494453849, 5.063625430476634, 4.903706519410671, 4.891005992072018, 4.857168214245934, 4.845526801893324, 4.845452341208768, 4.8205059750156, 4.799306005256147, 4.751...
scala> Hist.pointsPerBucket
// res15: Iterable[Int] = List(21, 23, 15, 22, 19)

我使用Lists而不是Arrays有点欺骗，但我希望你没关系

Answer 4

获取一些数字：

scala> val a = Array.tabulate(100){ _ => scala.util.Random.nextDouble * 10 }
a: Array[Double] = Array(6.333702962141718, 1.5506921990476974, 5.275795179538175, 1.7422259222209069, 2.8613172268857423, 9.489321343162988, 0.06102866084230496, 2.83927696305669...

使用groupBy将元素分组为＆＃34; size＆＃34; 2。

scala> a.groupBy{
     | case i if(i < 2) => "less than two"
     | case i if(i >= 2 && i < 4) => "two to four"
     | case i if(i >= 4 && i < 6) => "four to six"
     | case i if(i >= 6 && i < 8) => "six to eight"
     | case i if(i >= 8 && i <= 10) => "eight to ten"
     | }
res69: scala.collection.immutable.Map[String,Array[Double]] = Map(lessThanTwo -> Array(1.5506921990476974, 1.7422259222209069...

groupBy返回包含组的Map[String, Array[Double]]，现在我们可以遍历每个键的值并打印出符号以获得简单的直方图。

scala> res69.map(_._2).foreach(xs => {xs.foreach(xss => print("x")); println()})
xxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxx

Answer 5

另一个答案，我认为更简洁

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)

Answer 6

我有一个类似但略有不同的要求-根据用户定义的仓位/截止值制作直方图。假设，在OP的情况下，需要垃圾箱0-3，-4，-5，-6，-7,8 +。我尝试了几种方法，但是对我来说，突破是意识到我需要根据每个值放入其中的bin中的位置对数组进行分组：

unittest.TestResult class

在这种情况下的结果：

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

Scala简单直方图

6 个答案: