Scala 从一个范围到另一个范围采样

我想使用另一个

范围

对一个

范围

进行“采样”。例如：

def sample(in: Range, by: Range): Range = ???

// note: Range equals broken!
assert(sample(3 to 10     , 1 to 2     ).toVector == (4 to 5     ).toVector)
assert(sample(3 to 10     , 1 to 4 by 2).toVector == (4 to 6 by 2).toVector)
assert(sample(3 to 10 by 2, 1 to 2     ).toVector == (5 to 7 by 2).toVector)
assert(sample(3 to 10 by 2, 1 to 4 by 2).toVector == (5 to 9 by 4).toVector)

---

如何定义

示例

方法？

这里是一个尝试。我还没有测试所有可能的组合。为简单起见，

by

参数的起点必须大于或等于零，并且必须有一个正步长

def sample(in: Range, by: Range): Range = {
  val drop  = by.start
  val stepM = by.step
  require(drop >= 0 && stepM > 0)
  val in1 = in.drop(drop)
  val in2 = if (stepM == 1) 
    in1 
  else if (in1.isInclusive)  // copy-method is protected :(
    new Range.Inclusive(start = in1.start, end = in1.end, step = in1.step*stepM)
  else
    new Range          (start = in1.start, end = in1.end, step = in1.step*stepM)
  in2.take(by.size)
}

---

通过将

sample

返回类型更改为

IndexedSeq[Int]

，得到了这个简单的解决方案

def sample(in: Range, by: Range): IndexedSeq[Int] = {
  in intersect (in.start+by.start to in.start+by.end by by.step)
}

它搜索

in

和

by

之间的交集，并在

in.start

中移动

对于相应重构的断言，这两个

assert(sample(3 to 10     , 1 to 2     ) == (4 to 5     ))
assert(sample(3 to 10     , 1 to 4 by 2) == (4 to 6 by 2))

这两个都失败了

assert(sample(3 to 10 by 2, 1 to 2     ) == (5 to 7 by 2)) // intersect at 5 only
assert(sample(3 to 10 by 2, 1 to 4 by 2) == (5 to 9 by 4)) // empty intersect

---

我的尝试变得太混乱了，试图在所有情况下都是正确的。