问题描述
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）：