1、数组
1)、创建数组
创建定长数组和长度可变数组
//创建定长数组,使用new关键字,指定数组泛型为Int,长度为5,默认会初始化这5个值为指定类型的默认值,比如Int的话则5个默认值都是0
//注意:如果new,相当于调用了数组的apply方法,直接为数组赋值
scala> val arr = new Array[Int](5)
arr: Array[Int] = Array(0, 0, 0, 0, 0)
//修改数组元素的值
scala> arr(0)
res16: Int = 0
scala> arr(0) = 1
scala> arr(1) = 2
scala> arr(2) = 3
scala> arr
res20: Array[Int] = Array(1, 2, 3, 0, 0)
//直接创建定长数组并且初始化元素
scala> val arr2 = Array(1,2,3,4,5)
arr2: Array[Int] = Array(1, 2, 3, 4, 5)
scala> arr2
res21: Array[Int] = Array(1, 2, 3, 4, 5)
//打印数组,默认是打印hashcode,使用toBuffer则会打印数组元素
scala> println(arr)
[I@6c06b1bc
scala> println(arr.toBuffer)
ArrayBuffer(1, 2, 3, 0, 0)
//注意以下因为有没有使用new的区别
scala> val a = Array[Int](10)
a: Array[Int] = Array(10)
a.length = 1
scala> val b = new Array[Int](10)
b: Array[Int] = Array(0, 0, 0, 0, 0, 0, 0, 0, 0, 0)
a.length = 10
//创建变成数组(使用数组缓存ArrayBuffer)
scala> val ab = new ArrayBuffer[Int]()
//提示早不到ArrayBuffer类型,需要导包
:7: error: not found: type ArrayBuffer
val ab = new ArrayBuffer[Int]()
^
//按制表符键会给出提示
scala> import scala.collection.mutable.Array
ArrayBuffer ArrayBuilder ArrayLike ArrayOps ArraySeq ArrayStack
//导包完成
scala> import scala.collection.mutable.ArrayBuffer
import scala.collection.mutable.ArrayBuffer
//创建变长数组new ArrayBuffer[Int]()或者不需要new关键字
ArrayBuffer[Int](),既然是变长那么就不需要指定长度了
scala> val ab = new ArrayBuffer[Int]()
ab: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer()
//向数组中加入单个或多个元素 +=
scala> ab += 1
res24: ab.type = ArrayBuffer(1)
scala> ab += 2
res25: ab.type = ArrayBuffer(1, 2)
//加入多个元素
scala> ab += (3,4,5)
res26: ab.type = ArrayBuffer(1, 2, 3, 4, 5)
scala> ab
res27: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3, 4, 5)
2)、数组操作
//数组相加变长和定长的也可以相加 使用 ++=
//两个变长数组相加,ac中元素加入到ab中
scala> ab ++= ac
res32: ab.type = ArrayBuffer(1, 2, 3, 4, 5, 1, 2, 3, 4, 5)
scala> ab
res33: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3, 4, 5, 1, 2, 3, 4, 5)
//将定长数组元素加入到变长数组ac中
scala> ac ++= Array(6,7,8)
res34: ac.type = ArrayBuffer(1, 2, 3, 4, 5, 6, 7, 8)
scala> ac
res35: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3, 4, 5, 6, 7, 8)
//数组指定位置插入元素,使用insert
//在ac数组的角标为1的位置插入两个9
scala> ac.insert(1,9,9)
scala> ac
res38: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 9, 9, 2, 3, 4, 5, 6, 7, 8)
//ac角标为0的位置插入两个9
scala> ac.insert(0,9,9)
scala> ac
res40: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(9, 9, 1, 9, 9, 2, 3, 4, 5, 6, 7, 8)
//数组指定位置删除元素,使用remove
//ac的角标为0的位置删除2个元素
scala> ac.remove(0,2)
scala> ac
res42: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 9, 9, 2, 3, 4, 5, 6, 7, 8)
scala> ac.remove(1,2)
scala> ac
res44: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3, 4, 5, 6, 7, 8)
3)、遍历数组
使用for循环以及until
好用的until会生成脚标,0 until 10 包含0不包含10
scala> 0 until 10
res47: scala.collection.immutable.Range = Range(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
//变量数组
scala> for(i <- 0 until ac.length) print (ac(i) + " ")
1 2 3 4 5 6 7 8
4)、数组转换
yield关键字将原始的数组进行转换会产生一个新的数组,原始的数组不变
scala> val arr = Array(1,2,3,4,5,6,7,8,9)
arr: Array[Int] = Array(1, 2, 3, 4, 5, 6, 7, 8, 9)
//将arr中每个元素乘以10放到一个新的数组中
scala> val arr2 = for (i <- arr) yield i * 10
arr2: Array[Int] = Array(10, 20, 30, 40, 50, 60, 70, 80, 90)
scala> arr2
res0: Array[Int] = Array(10, 20, 30, 40, 50, 60, 70, 80, 90)
//而使用map更方便
scala> val arr3 = arr.map(_ * 10)
arr3: Array[Int] = Array(10, 20, 30, 40, 50, 60, 70, 80, 90)
scala> arr3
res2: Array[Int] = Array(10, 20, 30, 40, 50, 60, 70, 80, 90)
或者
scala> val arr4 = arr.map(x => x * 10)
arr4: Array[Int] = Array(10, 20, 30, 40, 50, 60, 70, 80, 90)
使用过滤器
scala> val arr2 = for( i <- arr if i % 2 == 0) yield i * 2
arr2: Array[Int] = Array(4, 8, 12, 16)
//对于上面这种情况直接使用map就不好使了,因为带有过滤条件,所以这里使用过滤器filter
scala> val arr5 = arr.filter(x => x % 2 ==0)
arr5: Array[Int] = Array(2, 4, 6, 8)
scala> val arr5 = arr.filter(x => x % 2 ==0).map(x => x * 2)
arr5: Array[Int] = Array(4, 8, 12, 16)
scala> arr5
res6: Array[Int] = Array(4, 8, 12, 16)
//或者使用_这种更简便的写法
scala> val arr6 = arr.filter(_ % 2 ==0).map(_ * 2)
arr6: Array[Int] = Array(4, 8, 12, 16)
scala> arr6
res7: Array[Int] = Array(4, 8, 12, 16)
5)、数组常用算法
//数组最大值、最小值、求和
scala> arr.sum
res18: Int = 41
scala> arr.max
res19: Int = 9
scala> arr.min
res20: Int = 1
//数组排序的一些方法
scala> val arr = Array(3,1,6,8,5,4,9,2,3)
arr: Array[Int] = Array(3, 1, 6, 8, 5, 4, 9, 2, 3)
scala> arr.sorted
res9: Array[Int] = Array(1, 2, 3, 3, 4, 5, 6, 8, 9)
scala> arr
res10: Array[Int] = Array(3, 1, 6, 8, 5, 4, 9, 2, 3)
scala> val arr2 = arr.sorted
arr2: Array[Int] = Array(1, 2, 3, 3, 4, 5, 6, 8, 9)
scala> arr2
res11: Array[Int] = Array(1, 2, 3, 3, 4, 5, 6, 8, 9)
scala> val arr2 = arr.sorted.reverse
arr2: Array[Int] = Array(9, 8, 6, 5, 4, 3, 3, 2, 1)
scala> arr2
res12: Array[Int] = Array(9, 8, 6, 5, 4, 3, 3, 2, 1)
scala> arr.sortBy(x => x)
res13: Array[Int] = Array(1, 2, 3, 3, 4, 5, 6, 8, 9)
scala> arr.sortWith(_>_)
res14: Array[Int] = Array(9, 8, 6, 5, 4, 3, 3, 2, 1)
scala> arr.sortWith(_<_)
res15: Array[Int] = Array(1, 2, 3, 3, 4, 5, 6, 8, 9)
scala> arr.sortWith((x,y)=> x arr.sortWith((x,y)=> x>y)
res17: Array[Int] = Array(9, 8, 6, 5, 4, 3, 3, 2, 1)
2、映射
在Scala中,把哈希表这种数据结构叫做映射
1)、创建映射
//第一种创建方式:使用"->" a -> b中好比a是键b是值
scala> val props = Map(1 -> 23,2 -> 45)
//用java的话来说就是键和值类型都是Int
props: scala.collection.immutable.Map[Int,Int] = Map(1 -> 23, 2 -> 45)
scala> val props = Map("user" -> "admin", "pwd" -> "1234")
//用java的话来说就是键和值类型都是String
props: scala.collection.immutable.Map[String,String] = Map(user -> admin, pwd -> 1234)
//第二种创建方式:使用(k,v)
scala> val map = Map((1,"tom"),(2,"jack"))
map: scala.collection.immutable.Map[Int,String] = Map(1 -> tom, 2 -> jack)
scala> val name = map(1)
name: String = tom
2、获取和修改映射中的值
//根据键取值
scala> val name = map(1)
name: String = tom
//修改值发现不行,因为上面的Map是不可变的,需要重新导包
scala> map(1) = "lucy"
:9: error: value update is not a member of scala.collection.immutable.Map[Int,String]
map(1) = "lucy"
//重新导入值可变的map包就能修改元素成功了
scala> import scala.collection.mutable.Map
Map MapBuilder MapLike MapProxy
scala> import scala.collection.mutable.Map
import scala.collection.mutable.Map
scala> val map = Map((1,"tom"),(2,"jack"))
map: scala.collection.mutable.Map[Int,String] = Map(2 -> jack, 1 -> tom)
scala> map(1) = "lucy"
scala> map
res25: scala.collection.mutable.Map[Int,String] = Map(2 -> jack, 1 -> lucy)
//往映射中添加元素
scala> map += ((3,"wangw"),(4,"lining"))
res26: map.type = Map(2 -> jack, 4 -> lining, 1 -> lucy, 3 -> wangw)
scala> map += (5 -> "xiaox",6 -> "jiej")
res27: map.type = Map(2 -> jack, 5 -> xiaox, 4 -> lining, 1 -> lucy, 3 -> wangw, 6 -> jiej)
//加的时候如果k一样那么会覆盖v,发5对应的xiaox被修改成了xiao
scala> map += (5 -> "xiao")
res28: map.type = Map(2 -> jack, 5 -> xiao, 4 -> lining, 1 -> lucy, 3 -> wangw, 6 -> jiej)
//两个映射相加 ++=
scala> val map2 = Map((1,"laowang"),(2,"lisi"))
map2: scala.collection.mutable.Map[Int,String] = Map(2 -> lisi, 1 -> laowang)
scala> val map3 = Map(3 ->"jinbm")
map3: scala.collection.mutable.Map[Int,String] = Map(3 -> jinbm)
//相加的时候如果k一样则v会覆盖,map3中元素加到map2中
scala> map2 ++= map3
res33: map2.type = Map(2 -> lisi, 1 -> laowang, 3 -> jinbm)
scala> map2
res34: scala.collection.mutable.Map[Int,String] = Map(2 -> lisi, 1 -> laowang, 3 -> jinbm)
注:如果映射中根据k不能取得v那么会报错
scala> map2(4)
java.util.NoSuchElementException: key not found: 4
at scala.collection.MapLike$class.default(MapLike.scala:228)
at scala.collection.AbstractMap.default(Map.scala:58)
...
//使用getOrElse来解决,如果找不到这个k对应的v那么给它赋个默认值
scala> map2.getOrElse(4,0)
res36: Any = 0