List 如何实施';takeUntil';一份名单?
我想查找第一个List 如何实施';takeUntil';一份名单?,list,scala,List,Scala,我想查找第一个7之前和之前的所有项目: val list = List(1,4,5,2,3,5,5,7,8,9,2,7,4) 我的解决办法是: list.takeWhile(_ != 7) ::: List(7) 结果是: List(1, 4, 5, 2, 3, 5, 5, 7) 还有其他解决方案吗?这里有一种方法可以通过foldLeft实现,还有一种尾部递归版本可以缩短长列表 还有我在玩这个游戏时使用的测试 import scala.annotation.tailrec import o
7
之前和之前的所有项目:
val list = List(1,4,5,2,3,5,5,7,8,9,2,7,4)
我的解决办法是:
list.takeWhile(_ != 7) ::: List(7)
结果是:
List(1, 4, 5, 2, 3, 5, 5, 7)
还有其他解决方案吗?这里有一种方法可以通过foldLeft实现,还有一种尾部递归版本可以缩短长列表 还有我在玩这个游戏时使用的测试
import scala.annotation.tailrec
import org.scalatest.WordSpec
import org.scalatest.Matchers
object TakeUntilInclusiveSpec {
implicit class TakeUntilInclusiveFoldLeft[T](val list: List[T]) extends AnyVal {
def takeUntilInclusive(p: T => Boolean): List[T] =
list.foldLeft( (false, List[T]()) )({
case ((false, acc), x) => (p(x), x :: acc)
case (res @ (true, acc), _) => res
})._2.reverse
}
implicit class TakeUntilInclusiveTailRec[T](val list: List[T]) extends AnyVal {
def takeUntilInclusive(p: T => Boolean): List[T] = {
@tailrec
def loop(acc: List[T], subList: List[T]): List[T] = subList match {
case Nil => acc.reverse
case x :: xs if p(x) => (x :: acc).reverse
case x :: xs => loop(x :: acc, xs)
}
loop(List[T](), list)
}
}
}
class TakeUntilInclusiveSpec extends WordSpec with Matchers {
//import TakeUntilInclusiveSpec.TakeUntilInclusiveFoldLeft
import TakeUntilInclusiveSpec.TakeUntilInclusiveTailRec
val `return` = afterWord("return")
object lists {
val one = List(1)
val oneToTen = List(1, 2, 3, 4, 5, 7, 8, 9, 10)
val boat = List("boat")
val rowYourBoat = List("row", "your", "boat")
}
"TakeUntilInclusive" when afterWord("given") {
"an empty list" should `return` {
"an empty list" in {
List[Int]().takeUntilInclusive(_ == 7) shouldBe Nil
List[String]().takeUntilInclusive(_ == "") shouldBe Nil
}
}
"a list without the matching element" should `return` {
"an identical list" in {
lists.one.takeUntilInclusive(_ == 20) shouldBe lists.one
lists.oneToTen.takeUntilInclusive(_ == 20) shouldBe lists.oneToTen
lists.boat.takeUntilInclusive(_.startsWith("a")) shouldBe lists.boat
lists.rowYourBoat.takeUntilInclusive(_.startsWith("a")) shouldBe lists.rowYourBoat
}
}
"a list containing one instance of the matching element in the last index" should `return`
{
"an identical list" in {
lists.one.takeUntilInclusive(_ == 1) shouldBe lists.one
lists.oneToTen.takeUntilInclusive(_ == 10) shouldBe lists.oneToTen
lists.boat.takeUntilInclusive(_ == "boat") shouldBe lists.boat
lists.rowYourBoat.takeUntilInclusive(_ == "boat") shouldBe lists.rowYourBoat
}
}
"a list containing one instance of the matching element" should `return` {
"the elements of the original list, up to and including the match" in {
lists.one.takeUntilInclusive(_ == 1) shouldBe List(1)
lists.oneToTen.takeUntilInclusive(_ == 5) shouldBe List(1,2,3,4,5)
lists.boat.takeUntilInclusive(_ == "boat") shouldBe List("boat")
lists.rowYourBoat.takeUntilInclusive(_ == "your") shouldBe List("row", "your")
}
}
"a list containing multiple instances of the matching element" should `return` {
"the elements of the original list, up to and including only the first match" in {
lists.oneToTen.takeUntilInclusive(_ % 3 == 0) shouldBe List(1,2,3)
lists.rowYourBoat.takeUntilInclusive(_.length == 4) shouldBe List("row", "your")
}
}
}
}
您可以使用以下功能
def takeUntil(list: List[Int]): List[Int] = list match {
case x :: xs if (x != 7) => x :: takeUntil(xs)
case x :: xs if (x == 7) => List(x)
case Nil => Nil
}
val list = List(1,4,5,2,3,5,5,7,8,9,2,7,4)
takeUntil(list) //List(1,4,5,2,3,5,5,7)
尾部递归版本
def takeUntilRec(list: List[Int]): List[Int] = {
@annotation.tailrec
def trf(head: Int, tail: List[Int], res: List[Int]): List[Int] = head match {
case x if (x != 7 && tail != Nil) => trf(tail.head, tail.tail, x :: res)
case x => x :: res
}
trf(list.head, list.tail, Nil).reverse
}
不耐烦者一行:
更通用的版本: 它以任何谓词作为参数。使用
span
执行主要作业:
implicit class TakeUntilListWrapper[T](list: List[T]) {
def takeUntil(predicate: T => Boolean):List[T] = {
list.span(predicate) match {
case (head, tail) => head ::: tail.take(1)
}
}
}
println(List(1,2,3,4,5,6,7,8,9).takeUntil(_ != 7))
//List(1, 2, 3, 4, 5, 6, 7)
println(List(1,2,3,4,5,6,7,8,7,9).takeUntil(_ != 7))
//List(1, 2, 3, 4, 5, 6, 7)
println(List(1,2,3,4,5,6,7,7,7,8,9).takeUntil(_ != 7))
//List(1, 2, 3, 4, 5, 6, 7)
println(List(1,2,3,4,5,6,8,9).takeUntil(_ != 7))
//List(1, 2, 3, 4, 5, 6, 8, 9)
尾部递归版本。 只是为了说明替代方法,它并不比以前的解决方案更有效
implicit class TakeUntilListWrapper[T](list: List[T]) {
def takeUntil(predicate: T => Boolean): List[T] = {
def rec(tail:List[T], accum:List[T]):List[T] = tail match {
case Nil => accum.reverse
case h :: t => rec(if (predicate(h)) t else Nil, h :: accum)
}
rec(list, Nil)
}
}
从
scala.collection.List
借用takeWhile实现并对其进行一些更改:
def takeUntil[A](l: List[A], p: A => Boolean): List[A] = {
val b = new scala.collection.mutable.ListBuffer[A]
var these = l
while (!these.isEmpty && p(these.head)) {
b += these.head
these = these.tail
}
if(!these.isEmpty && !p(these.head)) b += these.head
b.toList
}
可能的方法是:
def takeUntil[A](list:List[A])(predicate: A => Boolean):List[A] =
if(list.isEmpty) Nil
else if(predicate(list.head)) list.head::takeUntil(list.tail)(predicate)
else List(list.head)
使用内置函数的一些方法:
val list = List(1, 4, 5, 2, 3, 5, 5, 7, 8, 9, 2, 7, 4)
//> list : List[Int] = List(1, 4, 5, 2, 3, 5, 5, 7, 8, 9, 2, 7, 4)
//Using takeWhile with dropWhile
list.takeWhile(_ != 7) ++ list.dropWhile(_ != 7).take(1)
//> res0: List[Int] = List(1, 4, 5, 2, 3, 5, 5, 7)
//Using take with segmentLength
list.take(list.segmentLength(_ != 7, 0) + 1)
//> res1: List[Int] = List(1, 4, 5, 2, 3, 5, 5, 7)
//Using take with indexOf
list.take(list.indexOf(7) + 1)
//> res2: List[Int] = List(1, 4, 5, 2, 3, 5, 5, 7)
val list=列表(1,4,5,2,3,5,5,7,8,9,2,7,4)
//>list:list[Int]=list(1,4,5,2,3,5,5,7,8,9,2,7,4)
//将takeWhile与dropWhile一起使用
list.takeWhile(!=7)+list.dropWhile(!=7).take(1)
//>res0:List[Int]=List(1,4,5,2,3,5,5,7)
//使用带分段长度的take
list.take(list.segmentLength(!=7,0)+1)
//>res1:List[Int]=List(1,4,5,2,3,5,5,7)
//使用带indexOf的take
list.take(list.indexOf(7)+1)
//>res2:List[Int]=List(1,4,5,2,3,5,5,7)
这里的许多解决方案不是很有效,因为它们探索整个列表,而不是提前停止。下面是一个使用内置函数的简短解决方案:
def takeUntil[T](c: Iterable[T], f: T => Boolean): Iterable[T] = {
val index = c.indexWhere(f)
if (index == -1) c else c.take(index + 1)
}
顺序改变了:(没关系,我的takeWhile和dropWhile被切换了。如果列表中不包含7
,这将添加一个-这可能是OP想要的,也可能不是OP想要的,但对我来说似乎有点可疑(特别是因为它对空列表没有这样做).FYI,使用:+
将元素附加到列表中是非常低效的。当前您的解决方案是O(N^2)。使用列表缓冲区替换累加器,或者使用:
(前置)和reverse
最终的结果。@Aivean这本应该只是一个快速的演示版本,但它一直困扰着我。所以这里有一个O(n)foldLeft解决方案,以及O(n)如果我必须在生产代码中执行类似操作,我实际上会使用尾部递归版本。7
不应该包括在内,根据标准库的定义:scala>List(1,4,5,2,3,5,5,7,8,9,2,7,4)。takeWhile(!=7)res0:List[Int]=List(1,4,5,2,3,5,5)。提示-考虑使用<代码>列表→FordRebug 。但是总体上,我觉得你的问题中的大部分都是最短、最冗长的方式。我之所以使用这个答案是因为我实际上希望<>代码> Stutabor直到——并且这会在列表中保存一个迭代。这个函数不是尾部递归的,所以它可能会被堆栈失败。对于长列表,该方法的值较低。仅供参考。仅供参考,这种方法不是尾部递归的(在长输入列表中可能失败)。
def takeUntil[T](c: Iterable[T], f: T => Boolean): Iterable[T] = {
val index = c.indexWhere(f)
if (index == -1) c else c.take(index + 1)
}