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从文件中读取的字符串中删除特定字符和所有整数。

目标是从通过扫描仪读取的字符串中清除所有非句子结尾字符和数字。删除这些字符和整数的原因是为了从读入的单词中生成准确的字数、句子数和音节数。

原贴的代码很粗糙，已经修复，下面我自己重新贴一下。

感谢您为此付出的所有时间和帮助！

public class Word {
    private int wordCount, sentenceCount, syllableCount;
    private int nums [] = {1,2,3,4,5,6,7,8,9,0};
    private char vowels [] = {'a', 'e', 'i', 'o', 'u', 'y'};;
    private char punctuation [] = {'!', '?','.', ';', ':','-','"','(', ')'};;

    public Word()
    {
        wordCount = 0;
        sentenceCount = 0;
        syllableCount =0;
    }

    public Word(String next)
    {
        if(next.length() > 1)
        {
            //
            // Remove punctuation that does not create sentences
            //
            for(int i = 5; i < punctuation.length; i++)
                for(int j = 0; j < next.length(); j++)
                    if(next.charAt(j) == punctuation[i])
                            next = next.replace(j, "");
            //
            // Counting Sentences
            //
            for(int i=0; i < 5; i++)
                if(punctuation[i] == next.charAt(next.length()-1))
                    sentenceCount++;
            //
            // Remove numbers for accurate word counting
            //
            for(int i = 0; i < nums.length; i++)
                for(int j = 0; j < next.length(); j++)
                    if(next.charAt(j) == nums[i])
                        next = next.replace(j, "");
            //
            // Counts all syllables
            //
            for(int i = 0; i < vowels.length; i++)
                for(int j = 0; j < next.length()-1; j++)
                    if(vowels[i] == next.charAt(j))
                        syllableCount++;
            System.out.println(next);
        }
    }
4

3 回答 3

0

你不能写

next.return(j,"");// you are replacing the j value(int) with ""

因为 .replace() 是仅用另一个字符串替换 String 的东西.. 甚至 char 也无法替换.. 所以你需要输入 cast ..

所以我尝试并编辑了您的代码..让我知道您有一些问题...

public class Word {
private int wordCount, sentenceCount, syllableCount;
private int nums [] = {1,2,3,4,5,6,7,8,9,0};
private char vowels [] = {'a', 'e', 'i', 'o', 'u', 'y'};
private char punctuation [] = {'!', '?','.', ';', ':','-','(', ')'};

public Word()
{
    wordCount = 0;
    sentenceCount = 0;
    syllableCount =0;
}

public Word(String next)
{
    System.out.println("The Given String is"+next);
    if(next.length() > 1)
    {
        int n=0;
        int a=0;
            if(next.charAt(0)=='"'){n=1;}
        if(next.charAt(next.length()-1)=='"'){a=1;}

        for(int y=0+n; y<next.length()-a;y++){
             if(next.charAt(y)=='"'){
                next=next.substring(0,y)+next.substring(y+1,next.length());
                                    }}
        for(int i=0;i<8;i++){
              char k=punctuation[i];
              String l= Character.toString(k);
              int j=next.indexOf(k);
              if (next.contains(l)) {next=next.replace(l,"");}}

        for(int i=0;i<10;i++){
              int k=nums[i];
              String q= Integer.toString(k);
              if (next.contains(q)){
                  next=next.replace(q, "");}}

System.out.println("The String After Editing is: "+next);

}}}

于 2013-08-31T13:44:01.150 回答
0 
public class Word {
    private int wordCount, sentenceCount, syllableCount;
    private char vowels [] = {'a','e','i','o','u','y','A','E','I','O','U','Y'};
    private char punctuation [] = {'!','?','.',';',':','-','"','(',')',','};

    public Word()
    {
        wordCount = 0;
        sentenceCount = 0;
        syllableCount = 0;
    }

    public Word(String next)
    {
        // Remove numbers
        next = next.replaceAll("[0-9]", "");
        // Skip over blank tokens
        if(next.contains(" ") || next.length() < 1)
                return;

        // String builder method used for removing all non-sentence ending
        // grammar marks.
        StringBuilder newNext = new StringBuilder(next);
        String tempNext;
        if(newNext.length() > 0)
        {
            for(int i = 5; i < punctuation.length; i++)
                for(int j = 0; j < newNext.length(); j++)
                    if(punctuation[i] == newNext.charAt(j))
                    {
                         newNext = newNext.deleteCharAt(j);
                    }
            tempNext = newNext.toString();
            next = tempNext;

            if(next.length() < 1)
                return;
            else
                wordCount++;
        }

        // Count sentences
        for(int i = 0; i < 5; i++)
            if(next.charAt(next.length()-1) == punctuation[i])
                sentenceCount++;

        // Counts all syllables
        if(next.length() > 2)
        {
            for(int i = 0; i < vowels.length; i++)
                for(int j = 0; j < next.length()-1; j++)
                    if(vowels[i] == next.charAt(j))
                        syllableCount++;
        }
        //System.out.println(next);
    }
于 2013-09-03T17:00:55.147 回答
-1

如果word是您的方法.. 那么它的声明无效...对于您创建的任何方法，您都需要有一个返回类型.. 即，

如果方法返回值（假设 int）那么它需要

public int word(String S){}

如果该方法没有返回任何值，那么它需要是

public void word(String S){}

如果要构造构造函数..构造函数名称和类名称需要相同..

public World(String S){}
于 2013-08-31T02:06:00.927 回答