Morsecode 的熵表

如何解决Morsecode 的熵表

大家好，我有一个问题……我们必须做一项作业，我们必须将普通文本代码转换为 Processing (Java) 中的莫尔斯电码。我们还必须计算熵和冗余。翻译工作正常，但我们无法弄清楚如何计算所有字符的概率来计算熵。

    import controlP5.*;
import java.util.Map;
// dot = one unit
// dash = three units
// space between parts of the same letter = one unit
// space between letter = three units
// space between words is seven units

float x= 50;
float y= 450;
float w= 500;
float h = 80;

ControlP5 cp5;

String input;
String convert_string;
String output;
String temp;
Textarea myTextarea;
Textarea addTextarea;

String[] alphabet = {
  "a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"," "
};
String[] morse = {
  "._","_...","_._.","_..",".",".._.","__.","....","..",".___","_._","._..","__","_.","___",".__.","__._","._.","...","_",".._","..._",".__","_.._","_.__","__..","/"
};

int countLetter(String str,char zeichen) {
  str = str.toLowerCase();
  int count = 0;

  for (int i = 0; i < str.length(); i++) {
    char currentLetter = str.charAt(i);
    if (currentLetter == zeichen)
      count++;
  }

  return count;
}

double getAverageCharLength(String str) {
  String words[] = str.split("/");
  int numWords = words.length;
  int totalCharacters = 0;
  for (int i = 0; i < numWords; i++) {
    totalCharacters = totalCharacters + words[i].length();
  }
  return totalCharacters/numWords;
}


void setup() {
  size(600,600);
  convert_string= "";
  temp = "";
  input = "";
  output = "";
  cp5 = new ControlP5(this);

  cp5.addTextfield("Input").setPosition(50,100).setSize(100,50).setAutoClear(false);
  cp5.addBang("Submit").setPosition(50,200).setSize(100,50);
  myTextarea = cp5.addTextarea("Output")
    .setPosition(200,100)
    .setSize(300,300)
    .setFont(createFont("arial",20))
    .setLineHeight(25)
    .setColor(color(255))
    .setColorBackground(color(255,100))
    .setColorForeground(color(255,100));
  ;
  addTextarea = cp5.addTextarea("AddOutput")
    .setPosition(200,450)
    .setSize(300,100)
    .setFont(createFont("arial",14))
    .setLineHeight(14)
    .setColor(color(255))
    .setColorBackground(color(255,100));
  ;
  
}



void draw() {
  background(0);
  
}

public void Submit() {
  input = cp5.get(Textfield.class,"Input").getText();
  convert_string = input.toLowerCase();
  float otput = output.length();
  float oput =0;

  for (int i=0; i<convert_string.length(); i++) {
    for (int j=0; j<alphabet.length; j++) {
      if (String.valueOf(convert_string.charAt(i)).equals(alphabet[j])) {
        output += morse[j]+' ';
        otput= otput-1;
      }
    }
  }
 
double entropie = 0;
  double wortlaenge = 0;
  double charProbability=0;
  float curCharCount= 0;
  float inputLength = input.length();
  float entcounter=0;

  // Anzahl der einzelnen Zeichen im Eingabestring
  HashMap<Character,Integer> map = new HashMap<Character,Integer>();
  for(int i = 0; i < input.length(); i++){
    char c = input.toLowerCase().charAt(i);
    Integer val = map.get(c);
    if(val != null){
      map.put(c,new Integer(val + 1));
    }else{
      map.put(c,1);
    }
  

  // iteriert über Zeichen der Eingabe und berechnet Entropie und Wortlänge
  for (HashMap.Entry<Character,Integer> me : map.entrySet()) {
    curCharCount = me.getValue();
    charProbability = curCharCount / input.length();

    // Wortlänge für aktuelles Zeichen aus Morsekode-Map holen
 // if (convert_string.get(me.getKey())) != null) {
   //   wortlaenge += charProbability * get(me.getKey()).length();
    }
  }
  //entropie=entropie/entcounter;
  oput= output.length()+otput; 
  wortlaenge += charProbability * oput;
  
      entropie = charProbability * (Math.log(charProbability) / Math.log(2));
      
  println(wortlaenge);
  println(charProbability);
  println(entropie);
  println(inputLength);
  }
  //entropie = -entropie;
//  double redundanz = wortlaenge - entropie; 
  
  
  
  
/*  println(output);
  temp= output;
  myTextarea.setText(output);
  output = "";
  addTextarea.setText("Durchschnittliche Wortlänge : "+ getAverageCharLength(temp) +"\n");
   println(countLetter(temp,'_'));
   println(countLetter(temp,'.'));
   println(countLetter(temp,'/'));
   */
//}

解决方法

应该如何计算熵和冗余？你有公式或定义的链接吗？这将使您的需求更加清晰。

“计算所有字符的概率以计算熵”是什么意思？

Stack Overflow 上有两个问题也与莫尔斯电码和处理相关，所以这些可能会有所帮助（尽管它们似乎主要是关于解码莫尔斯电码）：

如何解决Morsecode 的熵表

解决方法

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