3、scala数组

时间:2023-03-09 08:09:36
3、scala数组

一、Array 、Array Buffer

1、Array

在Scala中,Array代表的含义与Java中类似,也是长度不可改变的数组。

此外,由于Scala与Java都是运行在JVM中,双方可以互相调用,因此Scala数组的底层实际上是Java数组。例如字符串数组在底层就是Java的String[],
整数数组在底层就是JaVa的Int[]。 数组初始化后,长度就固定下来了,而且元素全部根据其类型初始化:
val a=new Array[Int](10)
val a=new Array[String](10)
可以直接使用Array()创建数组,元素类型自动推断:
val a=Array("hello","world")
a(0)= "hi" ######
scala> val a = new Array[Int](10)
a: Array[Int] = Array(0, 0, 0, 0, 0, 0, 0, 0, 0, 0) scala> a(0) = 1 #scala 用() 访问元素 scala> a
res57: Array[Int] = Array(1, 0, 0, 0, 0, 0, 0, 0, 0, 0) scala> a(1)
res58: Int = 0 ######
scala> val a = new Array[String](10)
a: Array[String] = Array(null, null, null, null, null, null, null, null, null, null) scala> a(1)
res60: String = null scala> a(1) = "leo" scala> a(1)
res62: String = leo ######
scala> val a = Array("hello", "word")
a: Array[String] = Array(hello, word) scala> a
res63: Array[String] = Array(hello, word) scala> a(0) = "Hi" scala> a
res66: Array[String] = Array(Hi, word) ######
scala> val a = Array("leo", 30)
a: Array[Any] = Array(leo, 30) #自动类型推断 scala> a.length
res67: Int = 2

2、Array Buffer

在Scala中,如果需要类似于Java中的ArrayList这种长度可变的集合类,则可以使用ArrayBuffer。

//如果不想每次都使用全限定名,则可以预先导入ArrayBuffer类
import scala.collection.mutable.ArrayBuffer //使用ArrayBufer()的方式可以创建一个空的ArrayBuffer
val b=ArrayBuffertlnty() //使用+=操作符,可以添加一个元素,或者多个元素
b += 1
b +=(2,3,4,5) /使用++=操作符,可以添加其他集合中的所有元素
b ++= Array(6,7,8,9,10) //使用trimEnd()函数,可以从尾部截断指定个数的元素
b.trimEnd(5) ######
scala> import scala.collection.mutable.ArrayBuffer
import scala.collection.mutable.ArrayBuffer scala> val b = ArrayBuffer[Int]()
b: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer() scala> b += 1 #scala特有语法,往数组中加元素
res68: b.type = ArrayBuffer(1) scala> b += (2,3,4,5)
res69: b.type = ArrayBuffer(1, 2, 3, 4, 5) scala> b ++= Array(6,7,8,9,10)
res70: b.type = ArrayBuffer(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) scala> b.trimEnd(5) scala> b
res72: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3, 4, 5) ######
/使用insert()函数可以在指定位置插入元素; /但是这种操作效率很低,因为需要移动指定位置后的所有元素
b.insert(5,6)
b.insert(6,7,8,9,10) /使用remove()函数可以移除指定位置的元素
b.remove(1)
b.remove(1,3) //Array与ArrayBuffer可以互相进行转换
b.toArray
a.toBuffer ######
scala> b
res73: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3, 4, 5) scala> b.insert(5,6) scala> b
res75: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3, 4, 5, 6) scala> b.insert
insert insertAll scala> b.insert(6,7,8,9,10) scala> b
res77: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) scala> b.remove(1)
res78: Int = 2 scala> b
res79: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 3, 4, 5, 6, 7, 8, 9, 10) scala> b.remove(1,3) scala> b
res81: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 6, 7, 8, 9, 10) scala> var bArray = b.toArray
bArray: Array[Int] = Array(1, 6, 7, 8, 9, 10) scala> var aArrayBuffer = a.toBuffer
aArrayBuffer: scala.collection.mutable.Buffer[Any] = ArrayBuffer(leo, 30)

3、遍历Array ArrayBuffer

//使用for循环和until遍历Array/ArrayBufer

/使until是Richlnt提供的函数
for(i<-0 until b.length)
printin(b(i)) //跳跃遍历Array/ArrayBuffer
for(i <- 0 until(b.length,2))
printin(b(i) //从尾部遍历Array/ArrayBuffer
for(i<-(0 until b.length).reverse)
printin(bi)) /使用“增强for循环”遍历Array/ArrayBuffer
for(e <- b)
printin(e) ######
scala> for (i <- 0 until b.length) println(b(i))
1
6
7
8
9
10 scala> for (i <- 0 until (b.length, 2)) println(b(i)) #指定步长
1
7
9 scala> for (i <- (0 until b.length).reverse) println(b(i)) #反转
10
9
8
7
6
1 scala> for(e <- b) println(e) #增强for
1
6
7
8
9
10

4、数组常见操作

/数组元素求和
val a=Array(1,2,3,4,5)
val sum=a.sum /获取数组最大值
val max=a.max /对数组进行排序
scala.util.Sorting.quickSort(a) /获取数组中所有元素内容
a.mkString
a.mkString(",") #指定分隔符
a.mkString("<",",",">") //toString函数
a.toString
b.toString

二、数组转换

1、使用yield和函数式编程转换数组

//对Array进行转换,获取的还是Array
val a=Array(1,2,3,4,5)
val a2=for(ele <- a)yield ele * ele //对ArayBuffer进行转换,获取的还是ArrayBuffer
val b=ArrayBuffer[In]()
b +=(1,2,3,4,5)
val b2=for(ele <-b)yield ele * ele /结合if守卫,仅转换需要的元素
val a3 = for(ele <- if ele % 2 == 0)yield ele * ele /使用函数式编程转换数组(通常使用第一种方式)
a.filter (_%2 == 0).map(2 * _)
a.filter {_ % 2 == 0}map{2 * _} ##scala yield
对于for循环的每次迭代,yield都会生成一个将被记住的值。就像有一个你看不见的缓冲区,for循环的每一次迭代都会将另一个新的值添加到该缓冲区。 当for循环结束运行时,它将返回所有已赋值的集合。返回的集合的类型与迭代产生的类型相同,因此Map会生成Map,List将生成List,等等。 另外请注意,最初的集合没有改变。for / yield构造根据您指定的算法创建一个新的集合。 ######
scala> import scala.collection.mutable.ArrayBuffer
import scala.collection.mutable.ArrayBuffer scala> val b = ArrayBuffer[Int]()
b: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer() scala> b += (1,2,3,4,5)
res1: b.type = ArrayBuffer(1, 2, 3, 4, 5) scala> val b2 = for(ele <- b) yield ele * ele
b2: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 4, 9, 16, 25) scala> val a3 = for(ele <- a if ele % 2 == 0) yield ele * ele #yield
a3: Array[Int] = Array(4, 16) scala> a.filter(_ % 2 == 0).map(_ * 2) #
res2: Array[Int] = Array(4, 8) scala> a.filter { _ % 2 == 0} map{ _ * 2}
res3: Array[Int] = Array(4, 8)

2、算法案例

移除第一个负数之后的所有负数;

/构建数组
val a = ArrayBuffer[inty()
a +=(1,2,3,4,5,-1,-3,-5,-9) /每发现一个第一个负数之后的负数,就进行移除,性能较差,多次移动数组
var foundFirstNegative = false
var arrayLength = a.length
var index = 0
while(index < arrayLength){
if(a(index)>= 0){
index += 1
}else{
if(!foundFirstNegative){foundFirstNegative = true;index += 1}
else{a.remove(index);arrayLength -= 1}
}
} ############
//性能较高的

scala> import scala.collection.mutable.ArrayBuffer

import scala.collection.mutable.ArrayBuffer

scala> val a = ArrayBuffer[Int]()

a: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer()

scala> a += (1,2,3,4,5,-1,-3,-5,-9)

res0: a.type = ArrayBuffer(1, 2, 3, 4, 5, -1, -3, -5, -9)

scala> :paste

// Entering paste mode (ctrl-D to finish)

var fountFirstNegative = false

val keepIndexes = for (i <- 0 until a.length if !fountFirstNegative || a(i) >= 0

) yield {
   if (a(i) < 0) fountFirstNegative = true
   i

}

for (i <- 0 until keepIndexes.length) { a(i) = a(keepIndexes(i)) }

a.trimEnd(a.length - keepIndexes.length)

// Exiting paste mode, now interpreting.

fountFirstNegative: Boolean = true

keepIndexes: scala.collection.immutable.IndexedSeq[Int] = Vector(0, 1, 2, 3, 4, 5)

scala> a

res3: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3, 4, 5, -1)