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Scala basic tutorial -- 18 -- set (2)

2022-07-04 18:52:00 【Empty.】

Scala Basic course –18– aggregate （ Two ）

Chapter goal

  1. master Iterable Collection related content .
  2. master Seq Collection related content .
  3. master Set Collection related content .
  4. master Map Collection related content .
  5. Master the case of counting the number of characters .

1. Iterable

1.1 summary

Iterable Represents a set that can be iterated , It inherited Traversable Trait , It is also the parent of other sets . most important of all , It defines the get iterator (iterator) Methods : def iterator: Iterator[A], This is an abstract method , Its concrete implementation class needs to implement this method , So as to achieve The elements in the iterative return set .

1.2 classification

Traversable There are two ways to traverse data :

  • adopt iterator() Method realization , Iterate over the function of the element .

    This way belongs to Active iteration , We can go through hasNext() Check if there are any elements , And can actively call next() Method to get the element , namely : We can control the iteration process autonomously .

  • adopt foreach() Method realization , The function of traversing elements .

    This way belongs to Passive iteration , We only provide one function , It does not control the process of traversal , namely : The iterative process is controlled by the set itself .

1.3 Case a : Ergodic set

demand

  1. Define a list , Storage 1, 2, 3, 4, 5 These five numbers .
  2. adopt iterator() Method to traverse the above list .
  3. adopt foreach() Method to traverse the above list .

Reference code 

// Case study :  Traversing set elements .
object ClassDemo01 {
    
  def main(args: Array[String]): Unit = {
    
    //1.  Define a list ,  Storage 1, 2, 3, 4, 5 These five numbers .
    val list1 = (1 to 5).toList

    //2.  adopt iterator() Method to traverse the above list .
    val it: Iterator[Int] = list1.iterator
    while(it.hasNext) {
    
      val ele = it.next()
      println(ele)
      if(ele == 3) println(" Additional operations : " + ele * 10)
    }
    println("-" * 15)

    //3.  adopt foreach() Method to traverse the above list .
    list1.foreach(println(_))
  }
}

1.4 Case 2 : Group traversal

If you encounter take Iterable The elements in the object are divided into fixed size groups , Then traverse This demand , You can go through grouped() Method To implement the .

Format 

def grouped(size: Int): Iterator[Iterable[A]]

demand

  1. Define a Iterable aggregate , Storage 1~13 Between all the integers .
  2. adopt grouped() Method , Yes Iterable Collection in accordance with 5 Elements are grouped as a group , Traverse and print the results .

Reference code 

// Case study :  demonstration grouped Grouping function .
object ClassDemo02 {
    
  def main(args: Array[String]): Unit = {
    
    //1.  Definition Iterable aggregate ,  Storage 1~13 Number between .
    val i1 = (1 to 13).toIterable
    //2.  adopt grouped Method according to 5 A group of elements ,  Yes Iterable Set grouping .
    val result1 = i1.grouped(5)
    //3.  Traversing elements ,  Print the results .
    while (result1.hasNext) {
    
      println(result1.next())
    }
  }
}

1.5 Case three : Generate tuples according to the index

Iterable Every element stored in the collection is indexed , If we want to follow Elements -> Indexes This format , Generate a new set , here , It needs to be used zipWithIndex() Method 了 .

demand

  1. Define a Iterable aggregate , Storage "A", “B”, “C”, “D”, "E" These five strings .
  2. adopt zipWithIndex() Method , according to character string -> Index this format , Generate a new set .
  3. In line with Indexes -> String format , Generate a new set .
  4. Print the results .

Reference code 

// Case study :  Generate tuples according to the index 
object ClassDemo03 {
    
  def main(args: Array[String]): Unit = {
    
    //1.  Define a Iterable aggregate ,  Storage "A", "B", "C", "D", "E" These five strings .
    val i1 = Iterable("A", "B", "C", "D", "E")
    //2.  adopt zipWithIndex() Method ,  according to   character string -> Index this format ,  Generate a new set .
    val i2 = i1.zipWithIndex    //List((A,0), (B,1), (C,2))
    //3.  In line with   Indexes -> String format ,  Generate a new set .
    val i3 = i1.zipWithIndex.map(x => x._2 -> x._1)
    //4.  Print the results .
    println(i2)
    println(i3)
  }
}

1.6 Case four : Determine whether the set is the same

occasionally , We don't just want to Determine whether the elements in the two sets are all the same , It also requires that the iterative order of these two set elements be consistent , here , You can go through sameElements() Method To meet this requirement .

namely sameElements() The function of the method is : Judge whether the elements contained in the set and the iteration order of the elements are consistent .

demand

  1. Definition Iterable aggregate list1, contain "A", “B”, "C" These three elements .
  2. adopt sameElements() Method , Judge list1 and Iterable(“A”, “B”, “C”) Whether the set is the same .
  3. adopt sameElements() Method , Judge list1 and Iterable(“A”, “C”, “B”) Whether the set is the same .
  4. Definition Iterable aggregate list2, contain "A", “B”, “C”, "D" These four elements .
  5. adopt sameElements() Method , Judge list1 and list2 Are they the same? .

Reference code 

import scala.collection.mutable

// Case study :  Check two Iterable Whether it contains the same elements .
object ClassDemo04 {
    
  def main(args: Array[String]): Unit = {
    
    //1.  Definition Iterable aggregate i1,  contain "A", "B", "C" These three elements .
    val i1 = Iterable("A", "B", "C")
    //2.  Judge i1 and Iterable("A", "B", "C") Whether the set is the same .
    println(i1.sameElements(Iterable("A", "B", "C")))
    //3.  Judge i1 and Iterable("A", "C", "B") Whether the set is the same .
    println(i1.sameElements(Iterable("A", "C", "B")))
    //4.  Definition Iterable aggregate i2,  contain "A", "B", "C", "D" These four elements .
    val i2 = Iterable("A", "B", "C", "D")
    //5.  Judge i1 and i2 Are they the same? .
    println(i1.sameElements(i2))

    //6.  Expand :  establish HashSet Collective storage 1, 2,  establish TreeSet Collective storage 2, 1,  Then judge whether they are the same .
    val hs = mutable.HashSet(1, 2)
    val ts = mutable.TreeSet(2, 1)
    println(hs.sameElements(ts))    // The result is true,  because TreeSet Will sort the elements .
  }
}

2. Seq

2.1 summary

Seq Traits represent A sequence of elements arranged in a certain order , Sequence is a special iteratable set , Its element characteristics are Orderly ( The access order of elements is consistent ), repeatable , There is an index .

2.2 classification

The inheritance relationship is as follows :

Insert picture description here

IndexedSeq and LinearSeq yes Seq Sub characteristics of , It represents the two major sub sects of the sequence .

  1. IndexedSeq Trait Represents the index sequence , be relative to Seq Come on , It does not add additional methods , For random access to elements , It's more effective , Commonly used subsets are : NumericRange, Range, Vector, String etc. .

    • Range aggregate

      Range Represents an ordered integer queue , The spacing between each element is the same , For example, odd queue :1, 3, 5, 7, 9, But Fibonacci series 1, 1, 2, 3, 5, 8 It doesn't belong to Range class . Simply speaking , Range Similar to the arithmetic sequence in Mathematics .

    • NumericRange aggregate

      NumericRange Set is a more general arithmetic queue , Can generate Int, BigInteger, Short, Byte And other types of sequences .

    • Vector aggregate

      Vector Is a general immutable data structure , relatively , It takes a little longer to get data , But updating data randomly is much faster than arrays and linked lists .

  2. LinearSeq Trait Represents a linear sequence , It is realized through linked list , So it's head, tail, isEmpty It is relatively more efficient to implement . Commonly used subsets are : List, Stack, Queue etc. .

    • Stack: Express Stack data structure , The element features First in, then out .

      For historical reasons , Scala The current library also contains a immutable.Stack, But it is currently marked as deprecated , Because its design is not very elegant , And the performance is not very good , Because the stack will involve a large number of elements in and out , Therefore, the application scenario of immutable stack is relatively small , The most commonly used is variable stack , for example : mutable.Stack, mutable.ArrayStack.

      • mutable.Stack: It's through List, That is, the linked list , Additions and deletions quickly , Slow query .
      • mutable.ArrayStack: It's through Array, That is, the array method , Quick query , Add or delete slowly .

    • Queue: Express queue data structure , The element features fifo .

2.3 Case a : establish Seq aggregate

demand

establish Seq aggregate , Storage elements 1, 2, 3, 4, 5, And print the results .

Reference code 

// Case study :  establish Seq object .
object ClassDemo01 {
    
  def main(args: Array[String]): Unit = {
    
    //1.  establish Seq aggregate ,  Storage elements 1, 2, 3, 4, 5.
    val s1 = (1 to 5).toSeq
    //2.  Print the results .
    println(s1)
  }
}

2.4 Case 2 : Get the length and elements

because Seq Every element of the set is indexed , So we can get the corresponding elements directly through the index , And it can go through length() perhaps size() Method to get the length of the collection .

demand

  1. establish Seq aggregate , Storage elements 1, 2, 3, 4, 5.
  2. Print the length of the set .
  3. Get the index as 2 The elements of .

Reference code 

// Case study :  Get the length and elements 
object ClassDemo02 {
    
  def main(args: Array[String]): Unit = {
    
    //1.  establish Seq aggregate ,  Storage elements 1, 2, 3, 4, 5.
    val s1 = (1 to 5).toSeq
    //2.  Print the length of the set .
    println(s1.length, s1.size)
    println("-" * 15)

    //3.  Get the index as 2 The elements of .
    println(s1(2))
    println(s1.apply(2))
  }
}

2.5 Case three : Gets the index value of the specified element

Format

Gets the index value of the specified element , We can go through indexOf(), lastIndexOf(), indexWhere(), lastIndexWhere(),indexOfSlice() Method to implement , The specific functions are as follows :

  • indexOf: Get the position where the specified element first appears in the list .
  • lastIndexOf: Get the position of the last occurrence of the specified element in the list .
  • indexWhere: Get the element that meets the condition , The index that first appears in the set .
  • lastIndexWhere: Get the element that meets the condition , Index of the last occurrence in the set .
  • indexOfSlice: Gets the first occurrence position of the specified subsequence in the set .

Be careful : The above methods are to return the corresponding index after finding the specified data , If this data cannot be found , Then return to -1.

demand

  1. Definition Seq aggregate , Store the data : 1, 2, 4, 6, 4, 3, 2.
  2. Get elements 2 The index that first appears in the set , And print .
  3. Get elements 2 Index of the last occurrence in the set , And print .
  4. Get... In the collection , Less than 5 The first even index of , And print .
  5. From the index 2 The position starts to look in the set , Less than 5 The first even index of , And print .
  6. Get... In the collection , Less than 5 The last even index of , And print .
  7. Acquiring subsequence Seq(1, 2) stay t1 Collection , First occurrence index , And print .
  8. From the index 3 Start looking for subsequences Seq(1, 2) stay t1 Collection , First occurrence index , And print .

Reference code 

// Case study :  Gets the index value of the specified element 
object ClassDemo03 {
    
  def main(args: Array[String]): Unit = {
    
    //1.  Definition Seq aggregate ,  Store the data : 1, 2, 4, 6, 4, 3, 2.
    val s1 = Seq(1, 2, 4, 6, 4, 3, 2)
    //2.  Get elements 2 The index that first appears in the set ,  And print .
    println(s1.indexOf(2))
    //3.  Get elements 2 Index of the last occurrence in the set ,  And print .
    println(s1.lastIndexOf(2))
    println("-" * 15)

    //4.  Get... In the collection ,  Less than 5 The first even index of ,  And print .
    println(s1.indexWhere(x => (x % 2 == 0) && x < 5))
    //5.  From the index 2 The position starts to look in the set ,  Less than 5 The first even index of ,  And print .
    println(s1.indexWhere(x => (x % 2 == 0) && x < 5, 2))
    //6.  Get... In the collection ,  Less than 5 The last even index of ,  And print .
    println(s1.lastIndexWhere(x => (x % 2 == 0) && x < 5))
     //7.  Acquiring subsequence Seq(1, 2) stay t1 Collection ,  First occurrence index ,  And print .
    println(s1.indexOfSlice(Seq(1, 2)))
    //8.  From the index 3 Start looking for subsequences Seq(1, 2) stay t1 Collection ,  First occurrence index ,  And print .
    println(s1.indexOfSlice(Seq(1,2), 3))
  }
}

2.6 Case four : Determine whether the specified data is included

If we want to Determine whether the sequence begins or ends with the specified data , And judge whether the sequence contains the specified data , You can go through startsWith(), endsWith(), contains(), containsSlice() To implement the , As follows :

  • startsWith: Determine whether the set starts with the specified subsequence .
  • endsWith: Determine whether the set ends with the specified subsequence .
  • contains: Determine whether the set contains a specified data .
  • containsSlice: Determine whether the set contains a specified subsequence .

demand

  1. Definition Seq aggregate s1, Storage 1 To 10 These ten figures .
  2. Judge s1 Whether the set is in subsequence (1, 2) start , And print the results .
  3. Judge s1 Whether the set is in subsequence (1, 3) start , And print the results .
  4. Judge s1 Whether the set is in subsequence (9, 10) ending , And print the results .
  5. Judge s1 Whether the set is in subsequence (8, 9) ending , And print the results .
  6. Judge s1 Whether the collection contains elements 3, And print the results .
  7. Judge s1 Whether the set contains subsequences Seq(1, 2), And print the results .
  8. Judge s1 Whether the set contains subsequences Seq(1, 3), And print the results .

Reference code 

// Case study :  Determine whether the set contains the specified data 
object ClassDemo04 {
    
  def main(args: Array[String]): Unit = {
    
    //1.  Definition Seq aggregate s1,  Storage 1 To 10 These ten figures .
    val s1 = (1 to 10).toSeq
    //2.  Judge s1 Whether the set is in subsequence (1, 2) start ,  And print the results .
    println(s1.startsWith(Seq(1,2)))
    //3.  Judge s1 Whether the set is in subsequence (1, 3) start ,  And print the results .
    println(s1.startsWith(Seq(1,3)))
    //4.  Judge s1 Whether the set is in subsequence (9, 10) ending ,  And print the results .
    println(s1.endsWith(Seq(9,10)))
    //5.  Judge s1 Whether the set is in subsequence (8, 9) ending ,  And print the results .
    println(s1.endsWith(Seq(8,9)))
    println("-" * 15)
    //6.  Judge s1 Whether the collection contains elements 3,  And print the results .
    println(s1.contains(3))
    //7.  Judge s1 Whether the set contains subsequences Seq(1, 2),  And print the results .
    println(s1.containsSlice(Seq(1, 2)))
    //8.  Judge s1 Whether the set contains subsequences Seq(1, 3),  And print the results .
    println(s1.containsSlice(Seq(1, 3)))
  }
}

2.7 Case 5 : Modify the specified element

If we want to Modify an element value , Or use the specified subsequence to replace a certain element in the set , You can go through updated(), patch() Method to achieve , As follows :

  • updated: Modify the element of the specified index position to the specified value .

  • patch: Modify the element of the specified interval to the specified value .

demand

  1. Definition Seq aggregate s1, Storage 1 To 5 These five data .
  2. Modify the index 2 The element value of position is : 10, And print the results .
  3. From the index 1 Start , Use subsequence Seq(10, 20) Replace 3 Elements , And print the results .

Reference code 

// Case study :  Modify the specified element 
object ClassDemo05 {
    
  def main(args: Array[String]): Unit = {
    
    //1.  Definition Seq aggregate s1,  Storage 1 To 5 These five data .
    val s1 = (1 to 5).toSeq
    //2.  Modify the index 2 The element value of position is : 10,  And print the results .
    val s2 = s1.updated(2, 10)
    println(s2)
    //3.  From the index 1 Start ,  Use subsequence Seq(10, 20) Replace 3 Elements ,  And print the results .
    val s3 = s1.patch(1, Seq(10, 20), 3)
    println(s3)
  }
}

3. Stack

3.1 summary

Express Stack data structure , The element features First in, then out . For historical reasons , Scala The current library also contains a immutable.Stack, But it is currently marked as deprecated , Because its design is not very elegant , And the performance is not very good , Because the stack will involve a large number of elements in and out , Therefore, the application scenario of immutable stack is relatively small , The most commonly used is variable stack , for example : mutable.Stack, mutable.ArrayStack.

3.2 The illustration

Put elements 1, 2, 3 Add to stack , The diagram is as follows :
Insert picture description here

3.3 Common methods

  • top: Get stack top element , But this element will not be removed from the top of the stack .
  • push: Indicates the stack operation , It is equivalent to pushing elements into the top of the stack
  • pop: Remove the top element , And return this element .
  • clear: Remove all elements from the collection .

Be careful :

  1. immutable.Stack There is a unique method in the collection pushAll(), Push multiple elements onto the stack .
  2. mutable.ArrayStack The methods unique to the collection are :
    • dup(duplicate abbreviation ): It can be used to copy stack top elements .
    • preserving: This method will execute an expression , Recover the stack after the expression is executed , namely : The content of the stack is consistent with that before the call .

3.4 Example 1 : demonstration Stack Variable stack

demand

  1. Define variable stack Stack, Storage 1 To 5 this 5 A digital .
  2. adopt top() Method to get the top element , And print .
  3. adopt push() Method to put the element 6 Add to the top of the stack , And print .
  4. adopt pushAll() Add... To the top of the stack Seq(11, 22, 33) Sequence , And print .
  5. adopt pop() Method to remove the top element , And print .
  6. adopt clear() Method to remove all elements in the collection .

Reference code 

import scala.collection.mutable

// Case study :  demonstration Stack Variable stack 
object ClassDemo06 {
    
  def main(args: Array[String]): Unit = {
    
    //1.  Define variable stack Stack,  Storage 1 To 5 this 5 A digital .
    val s1 = mutable.Stack(1, 2, 3, 4, 5)
    //2.  adopt top() Method to get the top element ,  And print .
    println(s1.top)
    println(s1.size)
    println("-" * 15)
    //3.  adopt push() Method to put the element 6 Add to the top of the stack ,  And print .
    s1.push(6)
    println(s1)
    println("-" * 15)
    //4.  adopt pushAll() Add... To the top of the stack Seq(11, 22, 33) Sequence ,  And print .
    s1.pushAll(Seq(11, 22, 33))
    println(s1)
    println("-" * 15)
    //5.  adopt top() Method to remove the top element ,  And print .
    println(s1.pop())
    println(s1.size)
    //6.  adopt clear() Method to remove all elements in the collection .
    s1.clear()
    println(s1)
  }
}

3.5 Case 2 : demonstration ArrayStack Variable stack

demand

  1. Define variable stack ArrayStack, Storage 1 To 5 this 5 A digital .
  2. adopt dup() Method to copy the top element , And print the results .
  3. adopt preserving() Method realization First empty the set elements , Then recover the cleared data in the collection , And print .

Reference code 

import scala.collection.mutable

// Case study :  demonstration ArrayStack Variable stack 
object ClassDemo07 {
    
  def main(args: Array[String]): Unit = {
    
    //1.  Define variable stack ArrayStack,  Storage 1 To 5 this 5 A digital .
    val s1 = mutable.ArrayStack(1, 2, 3, 4, 5)
    //2.  adopt dup() Method to copy the top element ,  And print the results .
    s1.dup()
    println(s1)
    println("-" * 15)
    //3.  adopt preserving() Method realization ` First empty the set elements ,  Then recover the cleared data in the collection `,  And print .
    s1.preserving({
    s1.clear(); println(" See if I've done it !")})
    println(s1)
  }
}

4. Queue

4.1 summary

Represents a queue , The element features fifo , Our common queue is Variable queue : mutable.Queue, It is internally based on MutableList Data structure .

4.2 The illustration

Put elements 1, 2, 3 Add to queue , The diagram is as follows :

Insert picture description here

4.3 Common methods

  • enqueue: Queue method , You can pass in zero to multiple elements .
  • dequeue: Outgoing queue , Remove an element .
  • dequeueAll: Remove all elements that meet the conditions .
  • dequeueFirst: Remove the first element that meets the condition

4.4 Case presentation

demand

  1. Define variable queues Queue, Storage 1 To 5 These five data .
  2. Add elements to the queue 6, And print .
  3. Add elements to the queue 7, 8, 9, And print .
  4. Remove the first element of the queue , And print the element .
  5. Remove the first odd number of the queue , And print the element .
  6. Remove all even numbers in the queue , And print all removed data .
  7. Print variable queues Queue, See the end result .

Reference code 

import scala.collection.mutable

// Case study :  demonstration Queue Sequence .
object ClassDemo08 {
    
  def main(args: Array[String]): Unit = {
    
    //1.  Define variable queues Queue,  Storage 1 To 5 These five data .
    val q1 = mutable.Queue(1, 2, 3, 4, 5)
    //2.  Add elements to the queue 6,  And print .
    q1.enqueue(6)
    //3.  Add elements to the queue 7, 8, 9,  And print .
    q1.enqueue(7, 8, 9)
    println(q1)
    println("-" * 15)
    //4.  Remove the first element of the queue ,  And print the element .
    println(q1.dequeue())
    //5.  Remove the first odd number of the queue ,  And print the element .
    println(q1.dequeueFirst(_ % 2 != 0))
    //6.  Remove all even numbers in the queue ,  And print all removed data .
    println( q1.dequeueAll(_ % 2 == 0))
    //7.  Print variable queues Queue,  See the end result .
    println(q1)
  }
}

5. Set

5.1 summary

Set The elements in the collection do not contain duplicate elements , Common subcategories are : SortedSet( Subclass is TreeSet), HashSet, , ListSet.

5.2 classification

The inheritance relationship is as follows :

Insert picture description here

  • HashSet: Is a prefix tree ( Also called : Dictionary tree ) In the form of , The element features only , disorder , Often used for statistics .
  • LinkedHashSet: The element features only , Orderly .
  • TreeSet: The element features only , Sort .

5.3 Example

demand

  1. establish SortedSet aggregate , Storage elements 1, 4, 3, 2, 5, Then print the set .
  2. establish HashSet aggregate , Storage elements 1, 4, 3, 2, 5, Then print the set .
  3. establish LinkedHashSet aggregate , , Storage elements 1, 4, 3, 2, 5, Then print the set .

Reference code 

import scala.collection.{
    SortedSet, mutable}

// Case study :  demonstration Set aggregate .
object ClassDemo09 {
    
  def main(args: Array[String]): Unit = {
    
    //1.  establish SortedSet aggregate ,  Storage elements 1, 4, 3, 2, 5,  Then print the set .
    val s1 = SortedSet(1, 4, 3, 2, 5)
    println(s1)
    println("-" * 15)
    //2.  establish HashSet aggregate ,  Storage elements 1, 4, 3, 2, 5,  Then print the set .
    val s2 = mutable.HashSet(1, 4, 3, 2, 5)
    println(s2)
    println("-" * 15)
    //3.  establish LinkedHashSet aggregate , ,  Storage elements 1, 4, 3, 2, 5,  Then print the set .
    val s3 = mutable.LinkedHashSet(1, 4, 3, 2, 5)
    println(s3)
  }
}

6. Map

6.1 summary

Map Represent mapping , It contains key value pairs (key-value) Set , Map The basic operation of type is similar to Set Operations on collections , Because it contains elements Entry It's a key value pair , therefore Map Provides some methods for key or value operations alone .

6.2 classification

The inheritance relationship is as follows :

Insert picture description here

6.3 Example

demand

  1. Definition Map aggregate , Store data as : “A” -> 1, “B” -> 2, “C” -> 3.

  2. Traverse Map aggregate .

  3. adopt filterKeys() Method , Get the output key as "B" This set of key values of the object , And print the results .

    filterKeys: Filter for keys , Get the key value pairs that meet the conditions according to the key .

Reference code 

// Case study :  demonstration Map aggregate .
object ClassDemo10 {
    
  def main(args: Array[String]): Unit = {
    
    //1.  Definition Map aggregate ,  Store data as : "A" -> 1, "B" -> 2, "C" -> 3.
    val m1 = Map("A" -> 1, "B" -> 2, "C" -> 3)
    //2.  Traverse Map aggregate .
    m1.foreach(println(_))
    println("-" * 15)
    //3.  adopt filterKeys() Method ,  Get the output key as "B" This set of key values of the object ,  And print the results .
    println(m1.filterKeys(_ == "B"))
  }
}

7. Case study : Count the number of characters

7.1 demand

  1. Prompt the user to enter a string , And receive .
  2. Count the number of occurrences of each character in the above string , And print the results to the console .

7.2 Purpose

A comprehensive investigation aggregate , Keyboard Entry Related knowledge .

7.3 step

  1. Prompt the user to enter a string , And receive .
  2. Definition Map aggregate , Used to store characters and their occurrence times . key : character , value : The number of times characters appear .
  3. Convert string to character array .
  4. Traversal character array , Get every character .
  5. If the character appears for the first time , Record the number as 1, If the character is repeated , Just count the number of times +1, Then re store .
  6. Ergodic set , View results .

7.4 Reference code 

import scala.collection.mutable
import scala.io.StdIn

// Case study :  Count the number of characters .
object ClassDemo11 {
    
  def main(args: Array[String]): Unit = {
    
    //1.  Prompt the user to enter a string ,  And receive .
    println(" Please enter a string : ")
    val s = StdIn.readLine()
    //2.  Definition Map aggregate ,  Used to store characters and their occurrence times .  key : character ,  value :  The number of times characters appear .
    val m1 = mutable.Map[Char,Int]()
    //3.  Convert string to character array .
    val chs = s.toCharArray
    //4.  Traversal character array ,  Get every character .
    for(k <- chs) {
    
      //5.  If the character appears for the first time ,  Record the number as 1,  If the character is repeated ,  Just count the number of times +1,  Then re store .
      m1 += (k -> (if(m1.contains(k)) m1.getOrElse(k, 1) + 1 else 1))
    }
    //6.  Ergodic set ,  View results .
    for((k, v) <- m1) {
    
      println(k, v)
    }
  }
}

Recurring , Just count the number of times +1, Then re store .
6. Ergodic set , View results .

7.4 Reference code 

import scala.collection.mutable
import scala.io.StdIn

// Case study :  Count the number of characters .
object ClassDemo11 {
    
  def main(args: Array[String]): Unit = {
    
    //1.  Prompt the user to enter a string ,  And receive .
    println(" Please enter a string : ")
    val s = StdIn.readLine()
    //2.  Definition Map aggregate ,  Used to store characters and their occurrence times .  key : character ,  value :  The number of times characters appear .
    val m1 = mutable.Map[Char,Int]()
    //3.  Convert string to character array .
    val chs = s.toCharArray
    //4.  Traversal character array ,  Get every character .
    for(k <- chs) {
    
      //5.  If the character appears for the first time ,  Record the number as 1,  If the character is repeated ,  Just count the number of times +1,  Then re store .
      m1 += (k -> (if(m1.contains(k)) m1.getOrElse(k, 1) + 1 else 1))
    }
    //6.  Ergodic set ,  View results .
    for((k, v) <- m1) {
    
      println(k, v)
    }
  }
}
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