是否有 Porter2 词干分析器的 java 实现

Question

你知道 Porter2 词干分析器的 Java 实现（或者用 Java 编写的更好的词干分析器）吗？我知道这里有一个Java版本的Porter（不是Porter2）：

http://tartarus。 org/~martin/PorterStemmer/java.txt

但在 http://tartarus.org/~ martin/PorterStemmer/ 作者提到 Porter 有点过时，建议使用 Porter2，可在

http://snowball.tartarus.org/algorithms/english/stemmer.html

但是，我的问题是这个 Porter2 是用 Snowball 编写的（我以前从未听说过，所以不要'我对此一无所知）。我正在寻找的是它的java版本。

谢谢。他将非常感谢您的帮助。

Answer 1

Snowball 算法可作为 Java 下载

并从 snowball.tartarus.org：

2002 年 2 月 - Richard 获得了 Java 支持
修改了雪球代码生成器
产生 Java 输出以及 ANSI
C 输出。这意味着纯Java
系统现在可以使用雪球
词干提取器。

这就是你想要的，对吗？

您可以像这样创建它的实例：

  Class stemClass = Class.forName("org.tartarus.snowball.ext." + lang + "Stemmer");
  stemmer = (SnowballProgram) stemClass.newInstance();
  stemmer.setCurrent("your_word");
  stemmer.stem();
  String your_stemmed_word = stemmer.getCurrent();  

Answer 2

/*

   Porter stemmer in Java. The original paper is in

       Porter, 1980, An algorithm for suffix stripping, Program, Vol. 14,
       no. 3, pp 130-137,

   See also http://www.tartarus.org/~martin/PorterStemmer

   History:

   Release 1

   Bug 1 (reported by Gonzalo Parra 16/10/99) fixed as marked below.
   The words 'aed', 'eed', 'oed' leave k at 'a' for step 3, and b[k-1]
   is then out outside the bounds of b.

   Release 2

   Similarly,

   Bug 2 (reported by Steve Dyrdahl 22/2/00) fixed as marked below.
   'ion' by itself leaves j = -1 in the test for 'ion' in step 5, and
   b[j] is then outside the bounds of b.

   Release 3

   Considerably revised 4/9/00 in the light of many helpful suggestions
   from Brian Goetz of Quiotix Corporation ([email protected]).

   Release 4

*/

import java.io.*;

/**
  * Stemmer, implementing the Porter Stemming Algorithm
  *
  * The Stemmer class transforms a word into its root form.  The input
  * word can be provided a character at time (by calling add()), or at once
  * by calling one of the various stem(something) methods.
  */

class Stemmer
{  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)
   {  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];
   }

   /**
    * 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()
   {  k = i - 1;
      if (k > 1) { step1(); step2(); step3(); step4(); step5(); step6(); }
      i_end = k+1; i = 0;
   }

   /** Test program for demonstrating the Stemmer.  It reads text from a
    * a list of files, stems each word, and writes the result to standard
    * output. Note that the word stemmed is expected to be in lower case:
    * forcing lower case must be done outside the Stemmer class.
    * Usage: Stemmer file-name file-name ...
    */
   public static void main(String[] args)
   {
      char[] w = new char[501];
      Stemmer s = new Stemmer();
      for (int i = 0; i < args.length; i++)
      try
      {
         FileInputStream in = new FileInputStream(args[i]);

         try
         { while(true)

           {  int ch = in.read();
              if (Character.isLetter((char) ch))
              {
                 int j = 0;
                 while(true)
                 {  ch = Character.toLowerCase((char) ch);
                    w[j] = (char) ch;
                    if (j < 500) j++;
                    ch = in.read();
                    if (!Character.isLetter((char) ch))
                    {
                       /* to test add(char ch) */
                       for (int c = 0; c < j; c++) s.add(w[c]);

                       /* or, to test add(char[] w, int j) */
                       /* s.add(w, j); */

                       s.stem();
                       {  String u;

                          /* and now, to test toString() : */
                          u = s.toString();

                          /* to test getResultBuffer(), getResultLength() : */
                          /* u = new String(s.getResultBuffer(), 0, s.getResultLength()); */

                          System.out.print(u);
                       }
                       break;
                    }
                 }
              }
              if (ch < 0) break;
              System.out.print((char)ch);
           }
         }
         catch (IOException e)
         {  System.out.println("error reading " + args[i]);
            break;
         }
      }
      catch (FileNotFoundException e)
      {  System.out.println("file " + args[i] + " not found");
         break;
      }
   }
}

Answer 3

它作为 MG4J 的一部分提供。

请参阅 EnglishStemmer 的文档，即Porter2。使用方法 processTerm(MutableString ms)

MG4J 还为您提供了其他词干分析器的 java 版本。请参阅 snowball包。所有这些词干分析器都可以独立使用。

Answer 4

也许不是直接答案，但许多 NLP 工具包中都有词干分析器 - 请参阅 http://en.wikipedia。 org/wiki/Natural_language_processing_toolkits。
这里有一个相关的问题Tokenizer，stop Word Removal，Stemming in Java有几个可能有用的答案。

我们使用 OpenNLP，它是用 Java 编写的，可以提供该功能。如果您使用英语工作，我认为词干分析器之间的差异不会很重要。

Answer 5

看起来像 Lucene 以一种或另一种形式集成一些词干算法。您可能会从包 org.apache.lucene.analysis。然而，我担心词干代码会被深度集成到分析组件中，从而使其提取变得相当困难......

Answer 6

以下链接包含 Snowball Stemmer API。它具有 Porter Stemmer2 实现。
http://preciselyconcise.com/apis_and_installations/snowball_stemmer.php

Answer 7

这是我制作的轻量级包装，它是易于重用 和 可在 Maven Central 上获取。