【英文分词】Stemming Segmentation,基于词干分词

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英文分词主要是将各种时态形式的单词映射会同一种形式便于计算机理解。本文将列举一种基于词干分词的算法,这种算法建立在对英文构词法的充分了解上,所以代码中对各种形式的词汇的构词进行了拆解。

代码来源:http://tartarus.org/martin/PorterStemmer/,提供了多种语言版本。

如下为JAVA语言版本

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 ...    */}
测试分词器只需要如下三行代码:

Stemmer stem = new Stemmer(); //建立Stemmer对象String string = "created";    //目标单词stem.add(string.toCharArray(), string.toCharArray().length);  //加入单词到分词器stem.stem();   //分词System.out.println(stem.toString());  //输出结果
你会发现,分词结果不是create,而是creat,这是因为英文分词的目标是将不同形式映射到同一形式,而不是原始形式。同一形式足够计算机识别,并且省略末尾的e大大降低的分词算法的难度,在实现上更可行。


当然英文分词还有其他很多方式,我不精通,所以只罗列了一种词干分词的算法。


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