scala-筑基篇-01-List操作
2016-10-06 本文已影响108人
hylexus
[TOC]
List简介
特性
- 不可变
- 递归结构
- 同构(同质)的：元素的类型必须一致
- 协变的:如果S是T的子类型,那么List[S]是List[T]的子类型,这点不同于java的泛型
- 空列表为List[Nothing],Nil
创建列表
- 所有的List都是由空列表
Nil和操作符::构造出来的,::表示从前端扩张列表. - 实际上,用形如List(e1,e2,...)的方式创建列表在底层也是使用
Nil和::构造出来的
scala> List(1,2,3)
res0: List[Int] = List(1, 2, 3)
scala> 1::Nil
res1: List[Int] = List(1)
scala> 2::res1
res2: List[Int] = List(2, 1)
scala> 3::res2
res3: List[Int] = List(3, 2, 1)
操作
list的基本操作
| method | DESC |
|---|---|
| head | List的第一个元素 |
| tail | 除了head之外的其他元素组成的List |
| isEmpty | List是否为空 |
| last | List的最后一个元素 |
| init | 除了last之外的其他元素组成的List |
scala> val l1=List(1,2,3,4,5)
l1: List[Int] = List(1, 2, 3, 4, 5)
scala> l1.tail
res4: List[Int] = List(2, 3, 4, 5)
scala> l1.head
res5: Int = 1
scala> l1.isEmpty
res6: Boolean = false
scala> l1.init
res7: List[Int] = List(1, 2, 3, 4)
scala> l1.last
res8: Int = 5
list类的一阶方法
连接
列表的链接操作符:::和扩展元素操作符::一样都是右结合的,即xs:::ys:::zs等价于xs:::(ys:::zs),不过两个操作数都是List
scala> List(1,2,3):::List(4,5,6)
res9: List[Int] = List(1, 2, 3, 4, 5, 6)
长度
内部定义:
def length: Int = {
var these = self
var len = 0
while (!these.isEmpty) {
len += 1
these = these.tail
}
len
}
所以,length方法是比较费时的
reverse
反转list:reverse ,该方法并不是在原地修改list，因为list是不可变的，所以会返回一个新的list
- drop和take可以理解为更为广义的tail和init操作
- take(n)返回列表的前n个元素，
if(n>list.length) return list - drop(n)返回除了take(n)之外的所有元素，
if(n>list.length) return Nil - splitAt(n)在位置n处拆分列表,返回位元组。等价于
(list.take(n),list.drop(n))
scala> val l1=Range(1,11).toList
l1: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
scala> l1.take(3)
res10: List[Int] = List(1, 2, 3)
scala> l1.drop(3)
res11: List[Int] = List(4, 5, 6, 7, 8, 9, 10)
scala> l1.splitAt(3)
res12: (List[Int], List[Int]) = (List(1, 2, 3),List(4, 5, 6, 7, 8, 9, 10))
apply | indices
scala> val l1=Range(1,11).toList
l1: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
scala> l1.apply(2)
res13: Int = 3
scala> l1(2)
res14: Int = 3
scala> l1.indices
res15: scala.collection.immutable.Range = Range(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
zip
不匹配的元素将被遗弃
scala> val l1=Range(1,4).toList
l1: List[Int] = List(1, 2, 3)
scala> val l2=List("a","b","c","d")
l2: List[String] = List(a, b, c, d)
scala> l1.zip(l2)
res16: List[(Int, String)] = List((1,a), (2,b), (3,c))
mkString
mkString([start,]seperator[,end])
scala> val l=Range(1,5).toList
l: List[Int] = List(1, 2, 3, 4)
scala> l.mkString("start","|","end")
res17: String = start1|2|3|4end
scala> l.mkString("|")
res18: String = 1|2|3|4
list类的高阶方法
foreach
遍历list并将传入的lambda作用于每个元素
内部实现：
@inline final override def foreach[U](f: A => U) {
var these = this
while (!these.isEmpty) {
f(these.head)
these = these.tail
}
}
遍历
val l1 = Range(1, 11).toList
l1.foreach(e => { print(e + " ") })
l1.foreach(print(_))
等价的java8操作
List<Integer> list = Stream.iterate(1, i -> i + 1).limit(10)
.collect(Collectors.toList());
list.forEach(e -> {
System.out.print(e + " ");
});
list.forEach(System.out::print);
求和
val l1 = Range(1, 11).toList
var sum = 0
l1.foreach(sum += _)
println(sum)
map
参数f:T=>R,将f作用于每个元素,并返回类型为R的新列表
构造新list,元素为原list的元素的2倍
l1.map(e => e * 2)
println(l1)
等价的java8代码
list.stream().map(e->e*2).collect(Collectors.toList());
flatMap
和map类型,但请注意返回类型
scala> val l=List("tom","cat","apache")
l: List[String] = List(tom, cat, apache)
scala> l.map(_.toList)
res19: List[List[Char]] = List(List(t, o, m), List(c, a, t), List(a, p, a, c, h, e))
scala> l.flatMap(_.toList)
res20: List[Char] = List(t, o, m, c, a, t, a, p, a, c, h, e)
filter
scala> val l=Range(1,11).toList
l: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
scala> l.filter(e=>(e%2)==0)
res22: List[Int] = List(2, 4, 6, 8, 10)
scala> l.filter(e=>(e&1)==0)
res24: List[Int] = List(2, 4, 6, 8, 10)
partition
返回二元组(符合条件的部分,不符合条件的部分)
scala> val l=Range(1,11).toList
l: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
scala> l.partition(e=>(e%2)==0)
res26: (List[Int], List[Int]) = (List(2, 4, 6, 8, 10),List(1, 3, 5, 7, 9))
find
返回第一个满足条件的
scala> l.find(e=>(e%2)==0)
res27: Option[Int] = Some(2)
takeWhile | dropWhile | span
- takeWhile(p):返回能够满足p的最长前缀组成的list
- dropWhile(p):返回list中除了takeWhile(n)的部分组成的列表
- span:类似于slitAt结合了take和drop,span结合了takeWhile和dropWhile的结果
scala> val l=List(1,2,3,-1,2,3)
l: List[Int] = List(1, 2, 3, -1, 2, 3)
scala> l.dropWhile(_>0)
res28: List[Int] = List(-1, 2, 3)
scala> l.takeWhile(_>0)
res30: List[Int] = List(1, 2, 3)
scala> l.span(_>0)
res31: (List[Int], List[Int]) = (List(1, 2, 3),List(-1, 2, 3))
论断(forall,exists)
scala> val l=List(1,2,3,-1,2,3)
l: List[Int] = List(1, 2, 3, -1, 2, 3)
scala> l.forall(_>0)
res33: Boolean = false
scala> l.exists(_>0)
res35: Boolean = true
折叠
- 1.foldLeft(/:)
内部实现:
//注意op第一个参数类型为seed的类型
override /*TraversableLike*/
def foldLeft[B](z: B)(@deprecatedName('f) op: (B, A) => B): B = {
var acc = z//z表示初始值,seed
var these = this
while (!these.isEmpty) {
//每次将op作用于当前seed和当前元素并将结果重新赋值于seed
acc = op(acc, these.head)
these = these.tail
}
//最终返回"累加"的结果
acc
}
- 2.foldRight(:\)
内部实现:
//反转后调用foldLeft
//注意op第二个参数类型为seed的类型
override def foldRight[B](z: B)(op: (A, B) => B): B =
reverse.foldLeft(z)((right, left) => op(left, right))
- 3.fold
内部实现:
//注意op的两个参数类型相同
def fold[A1 >: A](z: A1)(op: (A1, A1) => A1): A1 = foldLeft(z)(op)
求和
val l1 = Range(1, 11).toList
var sum = l1.foldLeft(0)((acc, e) => acc + e)
println(sum) //55
sum = l1./:(0)((acc, e) => acc + e)
println(sum) //55
sum = l1.foldRight(0)((e, acc) => acc + e)
println(sum) //55
sum = l1.fold(0)((acc, e) => acc + e)
println(sum) //55
sum = l1.foldLeft(0)(_ + _)
println(sum) //55
sum = l1./:(0)(_ + _)
println(sum) //55
sum = (0 /: l1)(_ + _) //相当于 l1./:(0)
//sum = (l1 /: 0)(_ + _) //错误,相当于0./:(l1)
println(sum) //55
乘积
var pro = 1
val l1 = Range(1, 11).toList
pro = l1.foldLeft(1)((acc, e) => acc * e)
println(pro) //3628800
pro = (1 /: l1)(_ * _)
println(pro) //3628800
排序
- 1.sortWith(p: (Int, Int) => Boolean)
val l1 = Range(1, 11).toList
var l2 = l1.sortWith((l, r) => l > r)
println(l2) //List(10, 9, 8, 7, 6, 5, 4, 3, 2, 1)
l2 = l1.sortWith(_ > _)
println(l2) //List(10, 9, 8, 7, 6, 5, 4, 3, 2, 1)
