产生斯卡拉懒“螺旋”

Question

任务： 用于2D阵列的给定位置产生位于半径周围的位置的列表。

例如：

input: (1, 1)
radius: 1
output: ( (0, 0), (1, 0), (2, 0), 
          (0, 1),         (2, 1),
          (0, 2), (1, 2), (2, 2) ).

我写的东西像

def getPositions(x:Int, y:Int, r:Int) = {
  for(radius <- 1 to r) yield {
    List(
      for (dx <- -radius to radius) yield Pair(x + dx, y - radius),
      for (dx <- -radius to radius) yield Pair(x + dx, y + radius),
      for (dy <- -radius to radius) yield Pair(x + radius, y + dy),
      for (dy <- -radius to radius) yield Pair(x - radius, y + dy)
    )
  }
}

在此代码getPositions不返回点的sequance，但点的济源Tuple4的sequance。 我怎样才能“连击” 4所生成代码上市？或者是有我的任务更简洁的解决方案？ （我很新的阶）。

P.S。 它实际上是对我的星际争霸机器人。

Answer 1

您必须汇整清单（两次），因此这将做到：

def getPositions(x:Int, y:Int, r:Int) = {
  for(radius <- 1 to r) yield {
    List(
      for (dx <- -radius to radius) yield Pair(x + dx, y - radius),
      for (dx <- -radius to radius) yield Pair(x + dx, y + radius),
      for (dy <- -radius to radius) yield Pair(x + radius, y + dy),
      for (dy <- -radius to radius) yield Pair(x - radius, y + dy)
    ).flatten
  }
}.flatten

这不是一个“懒惰”螺旋，虽然。

修改

那一个是懒：

def P(i:Int, j:Int) = { print("eval"); Pair(i,j) }

def lazyPositions(x:Int, y:Int, r:Int) = {
  (1 to r).toStream.flatMap{ radius =>

    (-radius to radius).toStream.map(dx => P(x + dx, y - radius)) #:::
    (-radius to radius).toStream.map(dx => P(x + dx, y + radius)) #:::
    (-radius to radius).toStream.map(dy => P(x + radius, y + dy)) #:::
    (-radius to radius).toStream.map(dy => P(x - radius, y + dy))
  }
}


print(lazyPositions(1,1,1).take(3).toList) # prints exactly three times ‘eval’.

我使用def P方法展现真实懒惰。每次，你会创建一个Pair，它被调用。在一个懒惰的解决方案，你只想要这个需求。

Answer 2

尝试这种情况：

object Spiral
{
    def
    getPositions(x: Int, y: Int, r: Int): Seq[(Int, Int)] = {
      for { radius <- 1 to r
            dx <- -radius to radius
            dy <- -radius to radius
            if dx != 0 || dy != 0
      } yield
          (x + dx, y + dy)
    }


    def
    main(args: Array[String]): Unit = {
        printf("getPositions(1, 1, 1): %s%n", getPositions(0, 0, 1).mkString("{ ", ", ", " }"))
    }
}

输出：

getPositions(1, 1, 1): { (-1,-1), (-1,0), (-1,1), (0,-1), (0,1), (1,-1), (1,0), (1,1) }

Answer 3

您可以直接形成你的范围，并使用flatMap和++为他们所做的列表联合起来，你可能想在一个圆形的方向也去：

def getPositions(x: Int, y: Int, r: Int) = {
  (1 to r) flatMap (radius => {
    val dx = -radius to radius
    val dy = -(radius-1) to (radius-1)
    dx.map(i => (x+i, y+radius)) ++ dy.map(i => (x+radius, y-i)) ++
    dx.map(i => (x-i, y-radius)) ++ dy.map(i => (x-radius, y+i))
  })
}

如果你真的想要得到的结果是懒惰的，你必须做同样的懒惰组件：

def getPositions(x: Int, y: Int, r: Int) = {
  Stream.range(1,r+1) flatMap (radius => {
    val dx = Stream.range(-radius,radius+1)
    val dy = Stream.range(-(radius+1),radius)
    dx.map(i => (x+i, y+radius)) ++ dy.map(i => (x+radius, y-i)) ++
    dx.map(i => (x-i, y-radius)) ++ dy.map(i => (x-radius, y+i))
  })
}

编辑：固定了DX与DY错字

Answer 4

下面是这个问题的一些解决方案。首先，如果你不喜欢的顺序，只需位置，这样就可以了：

def getPositions(x:Int, y:Int, r:Int) = for {
  yr <- y - r to y + r
  xr <- x - r to x + r
  if xr != x || yr != y
} yield (xr, yr)

这会给你指定的完全相同的输出。不过，想要一个Python式发电机，所以这将是比较合适的：

def getPositions(x:Int, y:Int, r:Int) = Iterator.range(y - r, y + r + 1) flatMap {
  yr => Iterator.range(x - r, x + r + 1) map { 
    xr => (xr, yr)
  }
} filter (_ != (x, y))

这将返回一个Iterator，这可以通过使用next迭代。检查结束使用hasNext。

您可以替换Iterator或List或类似的东西Stream并获得完全生成集合。

现在，让我们假设你想有一个螺旋开始在中心和移动一次一个位置。我们可以像这样的东西做到这一点：

def getPositions(x:Int, y:Int, r:Int) = new Iterator[(Int, Int)] {
  private var currentX = x
  private var currentY = y
  private var currentR = 1
  private var incX = 0
  private var incY = 1
  def next = {
    currentX += incX
    currentY += incY
    val UpperLeft = (x - currentR, y + currentR)
    val UpperRight = (x + currentR, y + currentR)
    val LowerLeft = (x - currentR, y - currentR)
    val LowerRight = (x + currentR, y - currentR)
    val PrevSpiral = (x, y + currentR)
    val NextSpiral = (x - 1, y + currentR)
    (currentX, currentY) match {
      case NextSpiral => incX = 1; incY = 1; currentR += 1
      case PrevSpiral => incX = 1; incY = 0
      case UpperLeft => incX = 1; incY = 0
      case UpperRight => incX = 0; incY = -1
      case LowerRight => incX = -1; incY = 0
      case LowerLeft => incX = 0; incY = 1
      case _ =>
    }
    if (currentR > r)
      throw new NoSuchElementException("next on empty iterator")
    (currentX, currentY)
  }
  def hasNext = currentR <= r
}

Answer 5

下面是周围的边缘行走的流。

假定输入（3,3），2给出

{(1,1), (2,1), (3,1), (4,1), (5,1),
 (1,2),                      (5,2),
 (1,3),                      (5,3),
 (1,4),                      (5,4),
 (1,5), (2,5), (3,5), (4,5), (5,5)}

那么你可以使用以下内容：

def border(p: (Int,Int), r: Int) = {
  val X1 = p._1 - r
  val X2 = p._1 + r
  val Y1 = p._2 - r
  val Y2 = p._2 + r
  def stream(currentPoint: (Int,Int)): Stream[(Int,Int)] = {
    val nextPoint = currentPoint match {
      case (X1, Y1) => (X1+1, Y1)
      case (X2, Y2) => (X2-1, Y2)
      case (X1, Y2) => (X1, Y2-1)
      case (X2, Y1) => (X2, Y1+1)
      case (x, Y1) => (x+1, Y1)
      case (x, Y2) => (x-1, Y2)
      case (X1, y) => (X1, y-1)
      case (X2, y) => (X2, y+1)
    }
    Stream.cons(nextPoint, if (nextPoint == (X1,Y1)) Stream.empty else stream(nextPoint))
  }
  stream((X1,Y1))
}

用法：

scala> val b = border((3,3),2)
b: Stream[(Int, Int)] = Stream((2,1), ?)

scala> b.toList
res24: List[(Int, Int)] = List((2,1), (3,1), (4,1), (5,1), (5,2), (5,3), (5,4), (5,5), (4,5), (3,5), (2,5), (1,5), (1,4), (1,3), (1,2), (1,1))