Question

我必须编写一个程序，其中必须找到文件“ test.txt”中每个字母的多少，然后将该数字除以字母总数，以得出每个字母的出现百分比（我还没有完成）。但是，我什至不知道如何查找文本文件中每个字母的编号。这是我极其失败的尝试。谁能帮我。

import java.io.*;
class EnglishAnalysis
{
public static void main(String[] args)
{
try
{
  char letters[] = {'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'};
  FileReader fr = new FileReader("test.txt");
  BufferedReader br = new BufferedReader(fr);

  int count = 0;
  int totalLetters = 0;
  String lineCharacters [];
  String line;
  line = br.readLine();
  while (line != null)
  {
    lineCharacters = line.split("");
    for (int i = 0; i < lineCharacters.length; i++)
    {
      if (!Character.isWhitespace(line.charAt(i)))
      totalLetters++;
      for (int j = 0; j <= i; j++)
      {
        if (line.charAt(i) == letters[j])
          count++;
          System.out.println(line.charAt(i) + " : " + count);
      }
    }
    line = br.readLine();
  }
  System.out.println(totalLetters);
  br.close();
  }
catch (IOException e) {}
}
}

Answer 1

我采取了另一种方法。请参阅下面的代码和代码注释以获取解释。

代码：

package main;

import java.io.IOException;
import java.nio.charset.StandardCharsets;
import java.nio.file.Files;
import java.nio.file.Paths;
import java.text.DecimalFormat;

public class Main {

    public static void main(String[] args) {
        try {
            char letters[] = { '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' };
            double percentage[] = new double[26]; // declare a variable that will hold the percentages of each letter
            String path = "sample.txt";
            // Read all the contents of the file
            String text = new String(Files.readAllBytes(Paths.get(path)), StandardCharsets.UTF_8);
            // make a temporary copy of the content without all the letters
            String allLetters = text.replaceAll("[a-z]+", ""); //
            // Get the total number of characters in the file
            // by subtracting the new length (the length after all the letter has been
            // deleted) from the original length
            int totalNumOfLetters = text.length() - allLetters.length();

            // convert the contents into lower-case
            text = text.toLowerCase();
            for (int i = 0; i < letters.length; i++) {
                char c = letters[i];
                // make a temporary copy of the content without the current letter
                String tmp = text.replaceAll(c + "", "");

                // subtract the new length (the length after current letter has been deleted)
                // from the original length
                int count = text.length() - tmp.length();

                // get the percentage of the occurrence of the current letter
                percentage[i] = (double) count / (double) totalNumOfLetters * 100.00;
            }
            DecimalFormat df2 = new DecimalFormat(".##"); // for formatting only
            for (int i = 0; i < percentage.length; i++) {
                System.out.println("\"" + letters[i] + "\" has a percentage of " + df2.format(percentage[i]) + "%");
            }

        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
    }

}

示例文件：

'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'
'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h',

示例输出：

"a" has a percentage of 5.88%
"b" has a percentage of 5.88%
"c" has a percentage of 5.88%
"d" has a percentage of 5.88%
"e" has a percentage of 5.88%
"f" has a percentage of 5.88%
"g" has a percentage of 5.88%
"h" has a percentage of 5.88%
"i" has a percentage of 2.94%
"j" has a percentage of 2.94%
"k" has a percentage of 2.94%
"l" has a percentage of 2.94%
"m" has a percentage of 2.94%
"n" has a percentage of 2.94%
"o" has a percentage of 2.94%
"p" has a percentage of 2.94%
"q" has a percentage of 2.94%
"r" has a percentage of 2.94%
"s" has a percentage of 2.94%
"t" has a percentage of 2.94%
"u" has a percentage of 2.94%
"v" has a percentage of 2.94%
"w" has a percentage of 2.94%
"x" has a percentage of 2.94%
"y" has a percentage of 2.94%
"z" has a percentage of 2.94%

Answer 2

仅当该文件上的单词在一行上时，此方法才有效，但是可以使用空格。如果您仍然有问题，我明天也许可以解决此问题，但是明天我有课，所以我得睡觉了。希望这会有所帮助！

import java.io.*;
import java.util.*;

public class CountLettersFromFile {

public static void main(String[] args) throws IOException {

    FileReader file = new FileReader("test.txt");
    BufferedReader br=new BufferedReader(file);

    String[] lineCharacters;
    String line;

    line=br.readLine();
    int totalLetters=0;


    lineCharacters=line.split("");
    for (int i = 0; i < lineCharacters.length; i++) {
        if(!Character.isWhitespace(line.charAt(i)))
            totalLetters++;

    }


    double[] count = new double[255];
    int length = line.length();


    for (int i = 0; i < length; i++) {
        count[line.charAt(i)]++;
    }

    char[] ch = new char[line.length()];
    for (int i = 0; i < length; i++) {
        ch[i] = line.charAt(i);
        int find = 0;
        for (int j = 0; j <= i; j++) {
            if (line.charAt(i) == ch[j])
                find++;
        }

        if (find == 1 && !Character.isWhitespace(line.charAt(i))) {
            System.out.printf("Occurance of character \"" + line.charAt(i) + "\" is: %.0f %n"
                    , count[line.charAt(i)]);
            System.out.printf("Percent occurance of character \"" +line.charAt(i) + "\" is: %.3f %n"
                    , (count[line.charAt(i)])/totalLetters,"%");
            System.out.println("\n------------------------------------------------");
        }
    }
}

}

test.txt文件包括：

萨拉姆男子亚当男子亚当萨拉姆

输出：

字符“ s”的出现次数是：2

字符“ s”的出现百分比为：0.083

字符“ a”的出现是：8

字符“ a”的出现百分比是：0.333

字符“ l”的出现是：2

字符“ l”的出现百分比是：0.083

字符“ m”的出现是：6

字符“ m”的出现百分比是：0.250

字符“ e”的出现是：2

字符“ e”的出现百分比是：0.083

字符“ n”的出现是：2

字符“ n”的出现百分比是：0.083

字符“ d”的出现是：2

字符“ d”的出现百分比是：0.083

保留文本文件中字母出现的计数吗？

2 个答案: