Question

我想＆＃34;示例＆＃34; Range使用另一个Range。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sample方法？

Answer 1

这是一次尝试。我还没有测试过所有可能的组合。为简单起见，by参数的起点必须大于或等于零，并且是一个正步骤。

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)
}

Answer 2

通过将sample返回类型更改为IndexedSeq[Int]，获得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)
}

，in和by之间的交叉点的任务与in.start一起移位。

对于相应重构的断言，这两个传递，

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

用另一个范围对范围进行采样

2 个答案: