Spark基础-scala学习（集合）

Posted 2021-01-26 sky-chen

集合

1. scala的集合体系结构
2. List
3. LinkedList
4. Set
5. 集合的函数式编程
6. 函数式编程综合案例：统计多个文本内的单词总数

scala的集合体系结构

1. scala中的集合体系主要包括：Iterable、Seq、Set、Map。其中Iterable是所有集合trait的根trait。这个结构与java的集合体系非常相似
2. scala中的集合是分成可变和不可变两类集合的，其中可变集合就是说，集合的元素可以动态修改，而不可变集合的元素在初始化之后，就无法修改了。分别对应scala.collection.mutable和scala.collection.immutable两个包
3. Seq下包含了Range、ArrayBuffer、List等子trait。其中Range就代表了一个序列，通常可以使用“1 to 10”这种语法来产生一个Range。ArrayBuffer就类似于java中的ArrayList

List

1. List代表一个不可变的列表
2. List的创建，val list = List(1,2,3,4)
3. List有head和tail，head代表List的第一个元素，tail代表第一个元素之后的所有元素，list.head,list.tail
4. List有特殊的::操作符，可以用于将head和tail合并成一个List，0::list
5. 案例：用递归函数来给List中每个元素都加上指定前缀，并打印
6. 如果一个List只有一个元素，那么它的head就是这个元素，它的tail为Nil

scala> def decorator(l:List[Int],prefix:String){
     |  if(l != Nil){
     |   println(prefix+l.head)
     |   decorator(l.tail,prefix)
     |  }
     | }
decorator: (l: List[Int], prefix: String)Unit

scala> val list = List(1,2,3,5)
list: List[Int] = List(1, 2, 3, 5)

scala> decorator(list,"hello ")
hello 1
hello 2
hello 3
hello 5

scala> list.head
res1: Int = 1

scala> list.tail
res2: List[Int] = List(2, 3, 5)

scala> 8::list
res3: List[Int] = List(8, 1, 2, 3, 5)

LinkedList

1. LinkedList代表一个可变的列表，使用elem可以引用其头部，使用next可以引用其尾部
2. val l = scala.collection.mutable.LinkedList(1,2,3,4,5);l.elem;l.next
3. 案例：使用while循环while循环将列表中的每个元素都乘以2

scala> val list = scala.collection.mutable.LinkedList(1,2,3,5,6)

scala> var currentList = list
currentList: scala.collection.mutable.LinkedList[Int] = LinkedList(1, 2, 3, 5, 6)

scala> while(currentList != Nil){
     |  currentList.elem
     |  currentList.elem = currentList.elem * 2
     |  currentList = currentList.next
     | }

1. 案例：使用while循环将列表中每隔一个元素就乘以2

scala> :paste
// Entering paste mode (ctrl-D to finish)

val list = scala.collection.mutable.LinkedList(1,2,3,4,5,6,7,8,9,10)
var currentList = list
var first = true
while(currentList != Nil && currentList.next != Nil){
if(first){currentList.elem = currentList.elem * 2;first = false}
  currentList = currentList.next.next
  currentList.elem = currentList.elem * 2
  println(currentList.elem)
}

// Exiting paste mode, now interpreting.

<pastie>:11: warning: object LinkedList in package mutable is deprecated (since 2.11.0): low-level linked lists are deprecated
val list = scala.collection.mutable.LinkedList(1,2,3,4,5,6,7,8,9,10)
                                    ^
6
10
14
18
0
list: scala.collection.mutable.LinkedList[Int] = LinkedList(2, 2, 6, 4, 10, 6, 14, 8, 18, 10)
currentList: scala.collection.mutable.LinkedList[Int] = LinkedList()
first: Boolean = false

Set

1. Set代表一个没有重复元素的集合
2. 将重复元素加入Set是没有用的，比如val s = Set(1,2,3);s+1;s+4
3. 而且Set是不保证插入顺序的，也就是说，Set中的元素是乱序的，val s = new scala.collection.mutable.HashSetInt;s+=1;s+=2;s+=5
4. LinkedHashSet会用一个链表维护插入顺序，val s = new scala.collection.mutable.LinkedHashSetInt;i+=1;s+=2;s+=5
5. SrotedSet会自动根据key来进行排序，val s = scala.collection.mutable.SortedSet("orange","apple","banana")

scala> val s = Set(1,2,3)
s: scala.collection.immutable.Set[Int] = Set(1, 2, 3)

scala> s+1
res0: scala.collection.immutable.Set[Int] = Set(1, 2, 3)

scala> s+4
res1: scala.collection.immutable.Set[Int] = Set(1, 2, 3, 4)

scala> val s = new scala.collection.mutable.HashSet[Int]();s+=1;s+=2;s+=5
s: scala.collection.mutable.HashSet[Int] = Set(1, 5, 2)
res2: s.type = Set(1, 5, 2)

scala> val s = new scala.collection.mutable.LinkedHashSet[Int]();s+=1;s+=2;s+=5
s: scala.collection.mutable.LinkedHashSet[Int] = Set(1, 2, 5)
res4: s.type = Set(1, 2, 5)

scala> val s = scala.collection.mutable.SortedSet("orange","apple","banana")
s: scala.collection.mutable.SortedSet[String] = TreeSet(apple, banana, orange)

集合的函数式编程

scala> List("Leo","Jen","Peter","Jack").map("name is " + _)
res7: List[String] = List(name is Leo, name is Jen, name is Peter, name is Jack)

scala> List("Hello World","You Me").flatMap(_.split(" "))
res8: List[String] = List(Hello, World, You, Me)

scala> List("I","have","a","beautiful","house").foreach(println(_))
I
have
a
beautiful
house

scala> List("Leo","Jen","Peter","Jack").zip(List(100,90,75,83))
res10: List[(String, Int)] = List((Leo,100), (Jen,90), (Peter,75), (Jack,83))

综合案例统计多个文本内的单词总数

scala> val lines1 = lines01.mkString
lines1: String = /usr/bin/ruby -e "$(curl -fsSL https://raw.githubusercontent.com/Homebrew/install/master/install)"

scala> val lines2 = lines02.mkString
lines2: String = docker run -p 3307:3306 --name mysql3307 -v $PWD/conf:/etc/mysql/conf.d -v $PWD/logs:/logs -v $PWD/data:/var/lib/mysql -e MYSQL_ROOT_PASSWORD=123456 -d mysql:5.7

scala> val lines = List(lines1,lines2)
lines: List[String] = List(/usr/bin/ruby -e "$(curl -fsSL https://raw.githubusercontent.com/Homebrew/install/master/install)", docker run -p 3307:3306 --name mysql3307 -v $PWD/conf:/etc/mysql/conf.d -v $PWD/logs:/logs -v $PWD/data:/var/lib/mysql -e MYSQL_ROOT_PASSWORD=123456 -d mysql:5.7)

scala> lines.flatMap(_.split(" ")).map((_,1)).map(_._2).reduceLeft(_ + _)
res11: Int = 21

scala> lines.flatMap(_.split(" ")).map((_,1)).map(_._2)
res12: List[Int] = List(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1)

scala> lines.flatMap(_.split(" ")).map((_,1))
res13: List[(String, Int)] = List((/usr/bin/ruby,1), (-e,1), ("$(curl,1), (-fsSL,1), (https://raw.githubusercontent.com/Homebrew/install/master/install)",1), (docker,1), (run,1), (-p,1), (3307:3306,1), (--name,1), (mysql3307,1), (-v,1), ($PWD/conf:/etc/mysql/conf.d,1), (-v,1), ($PWD/logs:/logs,1), (-v,1), ($PWD/data:/var/lib/mysql,1), (-e,1), (MYSQL_ROOT_PASSWORD=123456,1), (-d,1), (mysql:5.7,1))