字母频率

Question

我有一个用户需要输入字符串的作业，我的程序到目前为止打印出它出现的次数，但我还需要显示字母的频率，例如，如果我输入“ab “ - 它会显示出来 字母发生频率   1 0.5   b 1 0.5

任何帮助都将深表感谢

提前致谢

import java.io.*;

public class ProgrammingAsignment {

public static void main (String [] args)throws IOException
{
    BufferedReader BR = new BufferedReader(new InputStreamReader(System.in));
    System.out.print("Enter Any Text:");
    String output = BR.readLine();
    output=output.toLowerCase();
    int length = output.length();
    char character;

    System.out.println("Letters\tFrequency\tCount");

    int count = 0;
    for(char i ='a'; i<='z'; i++)
    {
        count = 0;
        for(int j=0; j<length; j++)
        {
            character = output.charAt(j);
            if(character==i)
                count++;
        }
        if(count!=0)
        {
            System.out.println(i+"\t\t"+count);
        }
      }
    }

  }

Answer 1

您的代码现在打印出Occurs。因此，要执行频率，请在外部for循环中添加一个新变量，例如sum_count。在每个内部for循环迭代结束时，将count的值添加到sum_count，如下所示：

int sum_count = 0;
for(char i ='a'; i<='z'; i++)
{
        int count = 0;
        for(int j=0; j<length; j++)
        {
            character = output.charAt(j);
            if(character==i)
                count++;
        }
        sum_count += count; 
      ...
 }

现在，对于每个字母，只显示“已发生”的值，即count，并将其除以sum_count。

System.out.println(i+"\t\t"+count+"\t\t"+sum_count/count);

Answer 2

在数组中保存每个字母的计数。要获得任何字母的频率，请将其计数除以所有字母的计数总和。

Answer 3

最好的解决方案是构建一个大小为26的字符数组（从a到z的字符数）或52（如果你需要区分大小写）或128（如果你需要映射所有ascii字符） 。 例如：

int[] counter = new int[128];

在你的字符之间循环并为每个元素添加1。 例如，如果您找到A字母，则必须这样做：

counter[(int) 'A']++;

或者更通用的是，如果ch是当前的char：

counter[(int) ch]++; 

最后，您将获得计算频率所需的所有数据。

Answer 4

我会使用RegEx

   public CountFrequency wordString(String word, String input){
        String regex = "\\b" + word;
        Pattern p = Pattern.compile(regex);
        Matcher m = p.matcher(input); 
        int count = 0;

        while(m.find()){
         count++;
        }

        if(count =< 0){
            return new CountFrequency(0, 0.0);          
        }
        return new CountFrequency(count, (double)input.size()/count);       
   }




class CountFrequency{
        private int count = 0;
        private double frequency = 0.0;

        public CountFrequency(int count, double frequency){
            this.count = count;
            this.frequency = frequency;
        }

        int getCount(){
            return count;
        }

        double getFrequency(){
            return frequency;
        }
   }

Answer 5

这是一个使用地图存储遇到的字母并保持每个遇到的字母的最新数量的示例。地图允许您事先不知道在输出中将遇到多少个不同的字符。

package lettercount;
import java.util.*;
import java.io.*;

//Version that demonstrates using a map to keep a running count for items encountered
//This way you only have to step through the input once.
public class ProgrammingAssignment {

public static void main(String[] args) throws IOException {
    BufferedReader BR = new BufferedReader(new InputStreamReader(System.in));
    System.out.print("Enter Any Text: ");
    String output = BR.readLine();
    output=output.toLowerCase();

    int length = output.length();
    char character;
    int totalCount = 0;

    //we'll store each encountered character in this map, along with a count of the number
    //of times encountered.
    Map<Character, Integer> map = new HashMap<Character,Integer>();

    //Loop over the output once, character by character
    for (int i = 0; i < length; i++)
    {
        character = output.charAt(i);
        totalCount++; //This is the total number of characters we've found in the output

        Integer countForCharacter = 0;
        //check in map if we have a count for this character
        if (map.containsKey(character)) {
            //get the current count we have for this character
            countForCharacter = map.get(character);
            //increment
            countForCharacter++;
            //increment the count
        } else {
            countForCharacter = 1;
        }

        //Now put the up to date count into the map
        map.put(character, countForCharacter);
    }


    //Get the found characters as an array of Character
    Character[] charactersFound = map.keySet().toArray(new Character[0]);

    System.out.println("Letters\tFrequency\tCount");
    for(int k = 0; k < charactersFound.length; k++)
    {
        character = charactersFound[k];
        System.out.println(character+
                "\t" +
                //Following line gets the count for the character and divides by totalCount,
                //making sure that the the result is a floating point
                (map.get(character)/((float)totalCount)) +
                "\t"+
                //get the count for the character
                map.get(character));
    }
}

}