scala的多種集合的使用(5)之數組Array(ArrayBuffer)的操作

1.創建和更新數組的不同方式

1)定義一個數組的初始大小和類型，隨後填充值。

scala> val array = new Array[String](3)
array: Array[String] = Array(null, null, null)

scala> array(0) = "abc"

scala> array(1) = "def"

scala> array(2) = "hij"

scala> array
res66: Array[String] = Array(abc, def, hij)

2)創建指向數組的var引用，然後賦值給它。

scala> var fruits: Array[String] = _
fruits: Array[String] = null

scala> fruits = Array("apple","orange","banana")
fruits: Array[String] = [Ljava.lang.String;@79a73e5e

scala> fruits
res67: Array[String] = Array(apple, orange, banana)

3)集合混合類型組成。

scala> val x =Array(1,2.0,33D,400L)
x: Array[Double] = Array(1.0, 2.0, 33.0, 400.0)

4)集合混合類型組成，可以有自己控制。下面的例子的數組保持了原有數組的類型。

scala> val x =Array[Number](1,2.0,33D,400L)
x: Array[Number] = Array(1, 2.0, 33.0, 400)

5)range創建和填充數組。

scala> val x = Array.range(1,10)
x: Array[Int] = Array(1, 2, 3, 4, 5, 6, 7, 8, 9)

scala> val x = Array.range(1,10,2)
x: Array[Int] = Array(1, 3, 5, 7, 9)

6)fill創建和填充數組。

scala> val x = Array.fill(3)("hello")
x: Array[String] = Array(hello, hello, hello)

7)tabulate創建和填充數組。

scala> val x = Array.tabulate(5)(i => i * i)
x: Array[Int] = Array(0, 1, 4, 9, 16)

8)將list集合轉化爲數組的形式。

scala> val x = List(1,2,3,4,5).toArray
x: Array[Int] = Array(1, 2, 3, 4, 5)

9)將字符串轉化爲數組的形式。

scala> "hello".toArray
res68: Array[Char] = Array(h, e, l, l, o)

10)通過()的索引，就可以訪問對應的元素。

scala> val x = Array(1,3,4,5,7,9)
x: Array[Int] = Array(1, 3, 4, 5, 7, 9)

scala> x(0)
res70: Int = 1

scala> x(3)
res71: Int = 5

11)更新數組就是把訪問的元素給到一個值即可完成更新。

scala> x(3) = 30

scala> x
res73: Array[Int] = Array(1, 3, 4, 30, 7, 9)

2.創建大小可變的數組(ArrayBuffer)

1)Array是可變的，因爲它的內容可以改變，但是其大小是不能進行改變的。想要創建一個大小和值都可變的索引序列，可以使用ArrayBuffer。

scala> import scala.collection.mutable.ArrayBuffer
import scala.collection.mutable.ArrayBuffer

scala> var x = ArrayBuffer[String]()
x: scala.collection.mutable.ArrayBuffer[String] = ArrayBuffer()

scala> x += "aa"
res74: scala.collection.mutable.ArrayBuffer[String] = ArrayBuffer(aa)

scala> x += "bb"
res75: scala.collection.mutable.ArrayBuffer[String] = ArrayBuffer(aa, bb)

scala> x += "cc"
res76: scala.collection.mutable.ArrayBuffer[String] = ArrayBuffer(aa, bb, cc)

scala> x
res77: scala.collection.mutable.ArrayBuffer[String] = ArrayBuffer(aa, bb, cc)

2)可以在創建ArrayBuffer時就添加元素，之後繼續添加。

scala> val x = collection.mutable.ArrayBuffer("aa","bb")
x: scala.collection.mutable.ArrayBuffer[String] = ArrayBuffer(aa, bb)

scala> x += "cc"
res78: x.type = ArrayBuffer(aa, bb, cc)

3)可以使用++=添加元素到數組中去。

scala> val x = collection.mutable.ArrayBuffer("aa","bb")
x: scala.collection.mutable.ArrayBuffer[String] = ArrayBuffer(aa, bb)

scala> x ++= Seq("dd","ee")
res79: x.type = ArrayBuffer(aa, bb, dd, ee)

4)可以使append方法添加元素到數組中去。

scala> val x = collection.mutable.ArrayBuffer("aa","bb")
x: scala.collection.mutable.ArrayBuffer[String] = ArrayBuffer(aa, bb)

scala> x.append("ww","oo","pp")

scala> x
res81: scala.collection.mutable.ArrayBuffer[String] = ArrayBuffer(aa, bb, ww, oo, pp)

3.刪除Array和ArrayBuffer中的元素

 1)使用-=刪除一個或者多個元素。

scala> import scala.collection.mutable.ArrayBuffer
import scala.collection.mutable.ArrayBuffer

scala> val x = ArrayBuffer("c","d","a","e","f")
x: scala.collection.mutable.ArrayBuffer[String] = ArrayBuffer(c, d, a, e, f)

scala> x -= "a"
res82: x.type = ArrayBuffer(c, d, e, f)

scala> x -= ("c","d")
res83: x.type = ArrayBuffer(e, f)

2)使用--=刪除定義在另一個集合中的多個元素。

scala> val x = ArrayBuffer("c","d","a","e","f")
x: scala.collection.mutable.ArrayBuffer[String] = ArrayBuffer(c, d, a, e, f)

scala> x --= Seq("a","c")
res84: x.type = ArrayBuffer(d, e, f)

scala> x --= Array("d")
res85: x.type = ArrayBuffer(e, f)

scala> x --= Set("f")
res86: x.type = ArrayBuffer(e)

3)根據ArrayBuffer中元素的位置使用remove刪除元素，或者根據開始位置刪除一系列的元素。

scala> import scala.collection.mutable.ArrayBuffer
import scala.collection.mutable.ArrayBuffer

scala> val x = ArrayBuffer("c","d","a","e","f")
x: scala.collection.mutable.ArrayBuffer[String] = ArrayBuffer(c, d, a, e, f)

scala> x.remove(0)
res0: String = c

scala> x
res1: scala.collection.mutable.ArrayBuffer[String] = ArrayBuffer(d, a, e, f)

scala> x.remove(1,3)

scala> x
res3: scala.collection.mutable.ArrayBuffer[String] = ArrayBuffer(d)

4)clear方法可以刪除數組中的所有元素。

scala> import scala.collection.mutable.ArrayBuffer
import scala.collection.mutable.ArrayBuffer

scala> val x = ArrayBuffer("c","d","a","e","f")
x: scala.collection.mutable.ArrayBuffer[String] = ArrayBuffer(c, d, a, e, f)

scala> x.clear

scala> x
res5: scala.collection.mutable.ArrayBuffer[String] = ArrayBuffer()

5)定義數組變量var，通過過濾的方式刪除元素，重新將結果賦給自己。

scala> var a = Array("apple","orange","banana")
a: Array[String] = Array(apple, orange, banana)

scala> a = a.take(2)
a: Array[String] = [Ljava.lang.String;@405223e4

scala> a
res6: Array[String] = Array(apple, orange)

4.數組的排序

將一個數組或者ArrayBuffer中的元素進行排序。如果正在使用的是一個包含隱式排序元素的Array，可以使用scala.util.Sorting.quickSort方法對該Array進行排序。

scala> val x = Array("cherry","apple","banana")
x: Array[String] = Array(cherry, apple, banana)

scala> scala.util.Sorting.quickSort(x)

scala> x
res8: Array[String] = Array(apple, banana, cherry)

scala> val x = Array(2,4,1,9,5,7)
x: Array[Int] = Array(2, 4, 1, 9, 5, 7)

scala> scala.util.Sorting.quickSort(x)

scala> x
res10: Array[Int] = Array(1, 2, 4, 5, 7, 9)

5.創建多維數組

1)使用Array.ofDim方法創建所需要的數組。

scala> val rows = 2
rows: Int = 2

scala> val cols = 3
cols: Int = 3

scala> val x = Array.ofDim[String](rows,cols)

2)給多維數組添加元素。

scala> x(0)(0) = "a"

scala> x(0)(1) = "b"

scala> x(0)(2) = "c"

scala> x(1)(0) = "d"

scala> x(1)(1) = "e"

scala> x(1)(2) = "f"

scala> x
res17: Array[Array[String]] = Array(Array(a, b, c), Array(d, e, f))

3)根據索引的具體位置訪問元素。

scala> val a = x(1)(1)
a: String = e

用for循環遍歷數組的元素：

scala> for {
	 | i <- 0 until rows
	 | j <- 0 until cols
	 | } println(s"($i)($j) = ${x(i)(j)}")
(0)(0) = a
(0)(1) = b
(0)(2) = c
(1)(0) = d
(1)(1) = e
(1)(2) = f

4)創建三維數組。 

scala> val x,y,z = 3
x: Int = 3
y: Int = 3
z: Int = 3

scala> val a = Array.ofDim[Int](x,y,z)
a: Array[Array[Array[Int]]] = Array(Array(Array(0, 0, 0), Array(0, 0, 0), Array(0, 0, 0)), Array(Array(0, 0, 0), Array(0, 0, 0), Array(0, 0, 0)), Array(Array(0,0, 0), Array(0, 0, 0), Array(0, 0, 0)))

scala> for {
     | i <- 0 until x
     | j <- 0 until y
     | z <- 0 until z
     | } println(s"($i)($j)($z) = ${a(i)(j)(z)}")
(0)(0)(0) = 0
(0)(0)(1) = 0
(0)(0)(2) = 0
(0)(1)(0) = 0
(0)(1)(1) = 0
(0)(1)(2) = 0
(0)(2)(0) = 0
(0)(2)(1) = 0
(0)(2)(2) = 0
(1)(0)(0) = 0
(1)(0)(1) = 0
(1)(0)(2) = 0
(1)(1)(0) = 0
(1)(1)(1) = 0
(1)(1)(2) = 0
(1)(2)(0) = 0
(1)(2)(1) = 0
(1)(2)(2) = 0
(2)(0)(0) = 0
(2)(0)(1) = 0
(2)(0)(2) = 0
(2)(1)(0) = 0
(2)(1)(1) = 0
(2)(1)(2) = 0
(2)(2)(0) = 0
(2)(2)(1) = 0
(2)(2)(2) = 0