Scala函数式编程专题--scala集合和函数

//通过工厂，新建一个List，这个List是不可变的
scala> val numbers = List(1, 2, 3, 4, 5, 1, 2, 3, 4, 5)
numbers: List[Int] = List(1, 2, 3, 4, 5, 1, 2, 3, 4, 5)

scala> val tup = (1,1,2.1,"tuple",'c') //将多种不同数据结构打包一起，可以有重复
tup: (Int, Int, Double, String, Char) = (1,1,2.1,tuple,c)

scala> val tup = (0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21)
tup: (Int, Int, Int, Int, Int, Int, Int, Int, Int, Int, Int, Int, Int, Int, Int, Int, Int, Int, Int, Int, Int, Int) = (0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21)

//一个Tuple超过22个元素，报错了
scala> val tup = (0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22)
<console>:1: error: too many elements for tuple: 23, allowed: 22
val tup = (0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22)

//Option里面可以存普通数据类型
scala> val optionInt = Option(1)
optionInt: Option[Int] = Some(1)

scala> optionInt.get
res8: Int = 1
//但一个Option也可能为空
scala> val optionNone = Option(null)
optionNone: Option[Null] = None
//当Option里面是空的时候，是get不出东西的，还会报错
scala> optionNone.get
java.util.NoSuchElementException: None.get
 at scala.None$.get(Option.scala:347)
 at scala.None$.get(Option.scala:345)
 ... 32 elided
//这个时候可以判断它就是空的
scala> optionNone.isEmpty
res11: Boolean = true
//但可以用getOrElse()方法，当Option里面有东西的时候，就返回那个东西，如果没有东西，就返回getOrElse()的参数的内容
scala> optionNone.getOrElse("this is null") //里面没东西
res12: String = this is null
scala> optionInt.getOrElse("this is null") //里面有东西
res15: Any = 1

scala> val map = Map("test1" -> 1,"test2" -> 2)
map: scala.collection.immutable.Map[String,Int] = Map(test1 -> 1, test2 -> 2)

scala> map.get("test1")
res16: Option[Int] = Some(1)

scala> map.get("test3")
res17: Option[Int] = None

scala> map.get("test3").getOrElse("this is null")
res18: Any = this is null

//创建一个匿名函数
scala> val addOne = (x: Int) => x + 1
addOne: (Int) => Int = <function1>

scala> addOne(1)
res4: Int = 2

scala> val list = List(1,2,3)
list: List[Int] = List(1, 2, 3)

scala> list.map(_ + 1) //这里的(_+1)其实就是一个匿名函数 //让List中每一个元素+1，并返回
res29: List[Int] = List(2, 3, 4)

scala> list.reduce((i,j) => i + j) //两个两个一起循环，这里是让两个相加
res28: Int = 6

scala> val numbers = List(1, 2, 3, 4)
numbers: List[Int] = List(1, 2, 3, 4)

//过滤出余2等于0的
scala> numbers.filter((i: Int) => i % 2 == 0)
res0: List[Int] = List(2, 4)

scala> val numbers = List(1, 2, 3, 4)
numbers: List[Int] = List(1, 2, 3, 4)

//(m: Int, n: Int) => m + n这部分是一个匿名函数
scala> numbers.foldLeft(0)((m: Int, n: Int) => m + n)
res30: Int = 10

//下面的case就是一个偏函数PartialFunction[Int, String]
scala> val one: PartialFunction[Int, String] = { case 1 => "one" }
one: PartialFunction[Int,String] = <function1>

scala> one(1)
res11: String = Int

scala> one("one")
<console>:13: error: type mismatch;
 found : String("one")
 required: Int
 one("one")

//这个list里面的Any类型
scala> val list:List[Any] = List(1, 3, 5, "seven")
list: List[Any] = List(1, 3, 5, seven)

//使用map会报错，因为map接收的参数是普通函
scala> list.map { case i: Int => i + 1 }
scala.MatchError: seven (of class java.lang.String)
 at $anonfun$1.apply(<console>:13)
 at $anonfun$1.apply(<console>:13)
 at scala.collection.immutable.List.map(List.scala:277)
 ... 32 elided

 //但如果用collect函数就可以，因为collect接收的参数是偏函数，它会自动使用偏函数的一些特性，所以可以自动过滤掉不符合的数据类型
scala> list.collect { case i: Int => i + 1 }
res15: List[Int] = List(2, 4, 6)

//定义一个打印两个输出参数的函数
scala> def partial(i:Int,j:Int) : Unit = {
 | println(i)
 | println(j)
 | }
partial: (i: Int,j: Int)Unit

//赋一个值给上面那个函数，另一个参数不赋值，生成一个新的函数
scala> val partialFun = partial(5,_:Int)
partialFun: Int => Unit = <function1>

//只要一个参数就可以调用啦
scala> partialFun(10)
5
10

//我们知道函数可以这样定义，让它接收两个参数
scala> def curring(i:Int)(j: Int): Boolean = {false}
curring: (i: Int)(j: Int)Boolean

//可以把这个函数赋值给一个变量，注意变量的类型
scala> val curringVal:(Int => (Int => Boolean)) = curring _
curringVal: Int => (Int => Boolean) = <function1>

//可以让这个变量接收一个参数，又变成另一个函数了
scala> val curringVal_1 = curringVal(5)
curringVal_1: Int => Boolean = <function1>

//再用这个变量接收一个参数，终于能返回结果了
scala> curringVal_1(10)
res32: Boolean = false