如何在java中将char数组转换为字符串

时间:2013-10-10 11:29:39

标签: java

这是我的porter stemmer程序....我在stem()函数中得到char数组的输出....我试图将其转换为字符串...但它需要较旧的值而不是新的例如,“look”被定义为“look”.....“look”在char数组中(b [c] in stem())但是将其转换为string时取值“look” “
    包文件;

import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.FileReader;
import java.io.FileWriter;
import java.util.StringTokenizer;
import java.util.Vector;

/**
 *
 * @author sky
 */

public class stemmer {



    public static String line1,line,element,sentence,str;
  private char[] b;
   private int i,     /* offset into b */
               i_end, /* offset to end of stemmed word */
               j, k;
  private static final int INC = 50;
                     /* unit of size whereby b is increased */

   public stemmer()
   { 
        //b = new char[INC];
      i = 0;
      i_end = 0;
   }

   /**
    * Add a character to the word being stemmed.  When you are finished
    * adding characters, you can call stem(void) to stem the word.
    */

   public void add(char ch)
   {  
         System.out.println("in add() function");
       if (i == b.length)
      {  
          char[] new_b = new char[i+INC];
         for (int c = 0; c < i; c++)
         new_b[c] = b[c];
         b = new_b;
      }
      b[i++] = ch;

   }



   /** Adds wLen characters to the word being stemmed contained in a portion
    * of a char[] array. This is like repeated calls of add(char ch), but
    * faster.
    */

   public void add(char[] w, int wLen)
   {  if (i+wLen >= b.length)
      {  
         char[] new_b = new char[i+wLen+INC];
         for (int c = 0; c < i; c++) 
             new_b[c] = b[c];
             b = new_b;
      }
      for (int c = 0; c < wLen; c++)
          b[i++] = w[c];
   }
public void addstring(String s1)
{
  b=new  char[s1.length()];
  for(int k=0;k<s1.length();k++)
  {
         b[k] = s1.charAt(k);
  //System.out.println(b[k]);
  }  
  i=s1.length();
}
   /**
    * After a word has been stemmed, it can be retrieved by toString(),
    * or a reference to the internal buffer can be retrieved by getResultBuffer
    * and getResultLength (which is generally more efficient.)
    */
   public String toString() { return new String(b,0,i_end); }

   /**
    * Returns the length of the word resulting from the stemming process.
    */
   public int getResultLength() { return i_end; }

   /**
    * Returns a reference to a character buffer containing the results of
    * the stemming process.  You also need to consult getResultLength()
    * to determine the length of the result.
    */
   public char[] getResultBuffer() { return b; }

   /* cons(i) is true <=> b[i] is a consonant. */

   private final boolean cons(int i)
   {  switch (b[i])
      {  case 'a': case 'e': case 'i': case 'o': case 'u': return false;
         case 'y': return (i==0) ? true : !cons(i-1);
         default: return true;
      }
   }

   /* m() measures the number of consonant sequences between 0 and j. if c is
      a consonant sequence and v a vowel sequence, and <..> indicates arbitrary
      presence,

         <c><v>       gives 0
         <c>vc<v>     gives 1
         <c>vcvc<v>   gives 2
         <c>vcvcvc<v> gives 3
         ....
   */

   private final int m()
   {  int n = 0;
      int i = 0;
      while(true)
      {  if (i > j) return n;
         if (! cons(i)) break; i++;
      }
      i++;
      while(true)
      {  while(true)
         {  if (i > j) return n;
               if (cons(i)) break;
               i++;
         }
         i++;
         n++;
         while(true)
         {  if (i > j) return n;
            if (! cons(i)) break;
            i++;
         }
         i++;
       }
   }

   /* vowelinstem() is true <=> 0,...j contains a vowel */

   private final boolean vowelinstem()
   {  int i; for (i = 0; i <= j; i++) if (! cons(i)) return true;
      return false;
   }

   /* doublec(j) is true <=> j,(j-1) contain a double consonant. */

   private final boolean doublec(int j)
   {  if (j < 1) return false;
      if (b[j] != b[j-1]) return false;
      return cons(j);
   }

   /* cvc(i) is true <=> i-2,i-1,i has the form consonant - vowel - consonant
      and also if the second c is not w,x or y. this is used when trying to
      restore an e at the end of a short word. e.g.

         cav(e), lov(e), hop(e), crim(e), but
         snow, box, tray.

   */

   private final boolean cvc(int i)
   {  if (i < 2 || !cons(i) || cons(i-1) || !cons(i-2)) return false;
      {  int ch = b[i];
         if (ch == 'w' || ch == 'x' || ch == 'y') return false;
      }
      return true;
   }

   private final boolean ends(String s)
   { 

       int l = s.length();
   int o = k-l+1;
      if (o < 0) 
          return false;
      for (int i = 0; i < l; i++) 
          if (b[o+i] != s.charAt(i))
          return false;
      j = k-l;
      return true;
   }

   /* setto(s) sets (j+1),...k to the characters in the string s, readjusting
      k. */

   private final void setto(String s)
   {  int l = s.length();
      int o = j+1;
      for (int i = 0; i < l; i++) 
          b[o+i] = s.charAt(i);
      k = j+l;
   }

   /* r(s) is used further down. */

   private final void r(String s) { if (m() > 0) setto(s); }

   /* step1() gets rid of plurals and -ed or -ing. e.g.

          caresses  ->  caress
          ponies    ->  poni
          ties      ->  ti
          caress    ->  caress
          cats      ->  cat

          feed      ->  feed
          agreed    ->  agree
          disabled  ->  disable

          matting   ->  mat
          mating    ->  mate
          meeting   ->  meet
          milling   ->  mill
          messing   ->  mess

          meetings  ->  meet

   */

   private final void step1()
   { 

       if (b[k] == 's')
      {  if (ends("sses")) k -= 2; else
         if (ends("ies")) setto("i"); else
         if (b[k-1] != 's') k--;
      }
      if (ends("eed")) { if (m() > 0) k--; } else
      if ((ends("ed") || ends("ing")) && vowelinstem())
      {  k = j;
         if (ends("at")) setto("ate"); else
         if (ends("bl")) setto("ble"); else
         if (ends("iz")) setto("ize"); else
         if (doublec(k))
         {  k--;
            {  int ch = b[k];
               if (ch == 'l' || ch == 's' || ch == 'z') k++;
            }
         }
         else if (m() == 1 && cvc(k)) setto("e");
     }

   }

   /* step2() turns terminal y to i when there is another vowel in the stem. */

   private final void step2() { if (ends("y") && vowelinstem()) b[k] = 'i'; }

   /* step3() maps double suffices to single ones. so -ization ( = -ize plus
      -ation) maps to -ize etc. note that the string before the suffix must give
      m() > 0. */

   private final void step3() { if (k == 0) return; /* For Bug 1 */ switch (b[k-1])
   {
       case 'a': if (ends("ational")) { r("ate"); break; }
                 if (ends("tional")) { r("tion"); break; }
                 break;
       case 'c': if (ends("enci")) { r("ence"); break; }
                 if (ends("anci")) { r("ance"); break; }
                 break;
       case 'e': if (ends("izer")) { r("ize"); break; }
                 break;
       case 'l': if (ends("bli")) { r("ble"); break; }
                 if (ends("alli")) { r("al"); break; }
                 if (ends("entli")) { r("ent"); break; }
                 if (ends("eli")) { r("e"); break; }
                 if (ends("ousli")) { r("ous"); break; }
                 break;
       case 'o': if (ends("ization")) { r("ize"); break; }
                 if (ends("ation")) { r("ate"); break; }
                 if (ends("ator")) { r("ate"); break; }
                 break;
       case 's': if (ends("alism")) { r("al"); break; }
                 if (ends("iveness")) { r("ive"); break; }
                 if (ends("fulness")) { r("ful"); break; }
                 if (ends("ousness")) { r("ous"); break; }
                 break;
       case 't': if (ends("aliti")) { r("al"); break; }
                 if (ends("iviti")) { r("ive"); break; }
                 if (ends("biliti")) { r("ble"); break; }
                 break;
       case 'g': if (ends("logi")) { r("log"); break; }
   } }

   /* step4() deals with -ic-, -full, -ness etc. similar strategy to step3. */

   private final void step4() { switch (b[k])
   {
       case 'e': if (ends("icate")) { r("ic"); break; }
                 if (ends("ative")) { r(""); break; }
                 if (ends("alize")) { r("al"); break; }
                 break;
       case 'i': if (ends("iciti")) { r("ic"); break; }
                 break;
       case 'l': if (ends("ical")) { r("ic"); break; }
                 if (ends("ful")) { r(""); break; }
                 break;
       case 's': if (ends("ness")) { r(""); break; }
                 break;
   } }

   /* step5() takes off -ant, -ence etc., in context <c>vcvc<v>. */

   private final void step5()
   {   if (k == 0) return; /* for Bug 1 */ switch (b[k-1])
       {  case 'a': if (ends("al")) break; return;
          case 'c': if (ends("ance")) break;
                    if (ends("ence")) break; return;
          case 'e': if (ends("er")) break; return;
          case 'i': if (ends("ic")) break; return;
          case 'l': if (ends("able")) break;
                    if (ends("ible")) break; return;
          case 'n': if (ends("ant")) break;
                    if (ends("ement")) break;
                    if (ends("ment")) break;
                    /* element etc. not stripped before the m */
                    if (ends("ent")) break; return;
          case 'o': if (ends("ion") && j >= 0 && (b[j] == 's' || b[j] == 't')) break;
                                    /* j >= 0 fixes Bug 2 */
                    if (ends("ou")) break; return;
                    /* takes care of -ous */
          case 's': if (ends("ism")) break; return;
          case 't': if (ends("ate")) break;
                    if (ends("iti")) break; return;
          case 'u': if (ends("ous")) break; return;
          case 'v': if (ends("ive")) break; return;
          case 'z': if (ends("ize")) break; return;
          default: return;
       }
       if (m() > 1) k = j;
   }

   /* step6() removes a final -e if m() > 1. */

   private final void step6()
   {  j = k;
      if (b[k] == 'e')
      {  int a = m();
         if (a > 1 || a == 1 && !cvc(k-1)) k--;
      }
      if (b[k] == 'l' && doublec(k) && m() > 1) k--;
   }

   /** Stem the word placed into the Stemmer buffer through calls to add().
    * Returns true if the stemming process resulted in a word different
    * from the input.  You can retrieve the result with
    * getResultLength()/getResultBuffer() or toString().
    */
   public void stem()

   {  
       //System.out.println(i);
        k = i - 1;
      if (k > 1) 
      { 
          step1(); 
          step2();
          step3(); 
          step4(); 
          step5(); 
          step6(); 
      }
      for(int c=0;c<=k;c++)
      {
         // System.out.print(b[c]);

      }
      str=new String(b);
 //System.out.println(str);   
      sentence+=str+" ";
   System.out.println(sentence);  

      i_end = k+1; i = 0;
   }
   public static void main(String[] args)
   {
         stemmer s = new stemmer();

       try
       {
     BufferedReader  br = new BufferedReader(new FileReader("D:/output.txt"));
     BufferedWriter output = new BufferedWriter(new FileWriter("D:/output1.txt"));
     String separator = System.getProperty("line.separator");
       while ((line = br.readLine()) != null)
            {
                    StringTokenizer st2 = new StringTokenizer(line, "\n");

                    while (st2.hasMoreElements()) 
                    {
                        String line1 = (String) st2.nextElement();  
                        System.out.println(line1);
                        StringTokenizer st3 = new StringTokenizer(line1, " ");
                       // String sentence= new String();
                     while (st3.hasMoreTokens()) 
                     { 
                        element=st3.nextToken();
                        s.addstring(element);
                        s.stem();
                      }
     output.append( separator+sentence);
     sentence="";
                    }
            }
        output.close();
       }

     //System.out.println(element);   


//      s.addstring(element);
//      s.stem();
//      s.addstring("walks");
//      s.stem();
      //System.out.println("Output " +s.b);


       catch(Exception e)
       {

       }
}
}

1 个答案:

答案 0 :(得分:4)

要将char数组转换为String,请使用以下语句:

String str=new String(char[]);

请参阅docs