我试图为这个 SO question中提供的答案编写一个测试/计时函数。一些答案有效Array[T]
,一些有效List[T]
,一个有效Iterable[T]
,一个有效String
!
我想写的是一个函数,它shift*
从问题或答案、输入列表、谓词和预期输出中获取函数并运行该函数。有点像:
def test[T](
func:(Seq[T], T=>Boolean) => Seq[T],
input:Seq[T],
predicate:T=>Boolean,
expected:Seq[T]): Unit = {
// may be some warm up
// ... time start, run func, time stop,
// check output against expected
}
除了我可以弄清楚签名,因为Array
似乎具有可变Seq
属性,而List
似乎具有不可变Seq
属性。
处理它的最佳方法是什么?
编辑:使用 Thomas 的建议,这是我能得到的接近程度(适用于Array[Char]
,List[T]
但不适用于Array[T]
):
val inputArr = Array('a', 'b', 'C', 'D')
val expectArr = Array('a', 'C', 'D', 'b')
val inputList = inputArr.toList
val expectList = expectArr.toList
def test[I, T](
func:(I, T=>Boolean) => Traversable[T],
input: I,
predicate: T=>Boolean,
expected: Traversable[T]): Boolean = {
val result = func(input, predicate)
if (result.size == expected.size) {
result.toIterable.zip(expected.toIterable).forall(x => x._1 == x._2)
} else {
false
}
}
// this method is from Geoff [there][2]
def shiftElements[A](l: List[A], pred: A => Boolean): List[A] = {
def aux(lx: List[A], accum: List[A]): List[A] = {
lx match {
case Nil => accum
case a::b::xs if pred(b) && !pred(a) => aux(a::xs, b::accum)
case x::xs => aux(xs, x::accum)
}
}
aux(l, Nil).reverse
}
def shiftWithFor[T](a: Array[T], p: T => Boolean):Array[T] = {
for (i <- 0 until a.length - 1; if !p(a(i)) && p(a(i+1))) {
val tmp = a(i); a(i) = a(i+1); a(i+1) = tmp
}
a
}
def shiftWithFor2(a: Array[Char], p: Char => Boolean):Array[Char] = {
for (i <- 0 until a.length - 1; if !p(a(i)) && p(a(i+1))) {
val tmp = a(i); a(i) = a(i+1); a(i+1) = tmp
}
a
}
def shiftMe_?(c:Char): Boolean = c.isUpper
println(test(shiftElements[Char], inputList, shiftMe_?, expectList))
println(test(shiftWithFor2, inputArr, shiftMe_?, expectArr))
//following line does not compile
println(test(shiftWithFor, inputArr, shiftMe_?, expectArr))
//found : [T](Array[T], (T) => Boolean) => Array[T]
//required: (?, (?) => Boolean) => Traversable[?]
//following line does not compile
println(test(shiftWithFor[Char], inputArr, shiftMe_?, expectArr))
//found : => (Array[Char], (Char) => Boolean) => Array[Char]
//required: (?, (?) => Boolean) => Traversable[?]
//following line does not compile
println(test[Array[Char], Char](shiftWithFor[Char], inputArr, shiftMe_?, expectArr))
//found : => (Array[Char], (Char) => Boolean) => Array[Char]
//required: (Array[Char], (Char) => Boolean) => Traversable[Char]
我会将 Daniel 的答案标记为已接受,因为它编译并为我提供了一种不同的方式来实现我想要的 - 除非 Array[T] 上的方法创建一个新数组(并带来 Manifest 问题)。