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/*
* Licensed to the Apache Software Foundation (ASF) under one or more
* contributor license agreements. See the NOTICE file distributed with
* this work for additional information regarding copyright ownership.
* The ASF licenses this file to You under the Apache License, Version 2.0
* (the "License"); you may not use this file except in compliance with
* the License. You may obtain a copy of the License at
*
* http://www.apache.org/licenses/LICENSE-2.0
*
* Unless required by applicable law or agreed to in writing, software
* distributed under the License is distributed on an "AS IS" BASIS,
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
* limitations under the License.
*/
/*
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/index.html
Bug 1 (reported by Gonzalo Parra 16/10/99) fixed as marked below.
Tthe words 'aed', 'eed', 'oed' leave k at 'a' for step 3, and b[k-1]
is then out outside the bounds of b.
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.
[ This version is derived from Release 3, modified by Brian Goetz to
optimize for fewer object creations. ]
*/
package opennlp.tools.stemmer;
/**
*
* 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.
*/
// CHECKSTYLE:OFF
public class PorterStemmer implements Stemmer {
private char[] b;
private int i, /* offset into b */
j, k, k0;
private boolean dirty = false;
private static final int INC = 50;
public PorterStemmer() {
b = new char[INC];
i = 0;
}
/**
* reset() resets the stemmer so it can stem another word. If you invoke
* the stemmer by calling add(char) and then stem(), you must call reset()
* before starting another word.
*/
public void reset() { i = 0; dirty = false; }
/**
* Add a character to the word being stemmed. When you are finished
* adding characters, you can call stem(void) to process the word.
*/
public void add(char ch) {
if (b.length == i) {
char[] new_b = new char[i+INC];
System.arraycopy(b, 0, new_b, 0, i);
{
b = new_b;
}
}
b[i++] = ch;
}
/**
* 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.)
*/
@Override
public String toString() { return new String(b,0,i); }
/**
* Returns the length of the word resulting from the stemming process.
*/
public int getResultLength() { return i; }
/**
* 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 boolean cons(int i) {
switch (b[i]) {
case 'a': case 'e': case 'i': case 'o': case 'u':
return false;
case 'y':
return (i == k0) || !cons(i - 1);
default:
return true;
}
}
/* m() measures the number of consonant sequences between k0 and j. if c is
a consonant sequence and v a vowel sequence, and <..> indicates arbitrary
presence,
gives 0
vc gives 1
vcvc gives 2
vcvcvc gives 3
....
*/
private int m() {
int n = 0;
int i = k0;
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 <=> k0,...j contains a vowel */
private boolean vowelinstem() {
int i;
for (i = k0; i <= j; i++)
if (! cons(i))
return true;
return false;
}
/* doublec(j) is true <=> j,(j-1) contain a double consonant. */
private boolean doublec(int j) {
return j >= k0 + 1 && b[j] == b[j - 1] && 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 boolean cvc(int i) {
if (i < k0+2 || !cons(i) || cons(i-1) || !cons(i-2))
return false;
else {
int ch = b[i];
if (ch == 'w' || ch == 'x' || ch == 'y') return false;
}
return true;
}
private boolean ends(String s) {
int l = s.length();
int o = k-l+1;
if (o < k0)
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. */
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;
dirty = true;
}
/* r(s) is used further down. */
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 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)) {
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 void step2() {
if (ends("y") && vowelinstem()) {
b[k] = 'i';
dirty = true;
}
}
/* 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 void step3() {
if (k == k0) 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 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 vcvc. */
private void step5() {
if (k == k0) 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 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 a word provided as a String. Returns the result as a String.
*/
public String stem(String s) {
if (stem(s.toCharArray(), s.length()))
return toString();
else
return s;
}
/**
* Stem a word provided as a CharSequence.
* Returns the result as a CharSequence.
*/
public CharSequence stem(CharSequence word) {
return stem(word.toString());
}
/** Stem a word contained in a char[]. 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 boolean stem(char[] word) {
return stem(word, word.length);
}
/** Stem a word contained in a portion of a char[] array. 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 boolean stem(char[] wordBuffer, int offset, int wordLen) {
reset();
if (b.length < wordLen) {
b = new char[wordLen - offset];
}
System.arraycopy(wordBuffer, offset, b, 0, wordLen);
i = wordLen;
return stem(0);
}
/** Stem a word contained in a leading portion of a char[] array.
* 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 boolean stem(char[] word, int wordLen) {
return stem(word, 0, wordLen);
}
/** 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 boolean stem() {
return stem(0);
}
public boolean stem(int i0) {
k = i - 1;
k0 = i0;
if (k > k0+1) {
step1(); step2(); step3(); step4(); step5(); step6();
}
// Also, a word is considered dirty if we lopped off letters
// Thanks to Ifigenia Vairelles for pointing this out.
if (i != k+1)
dirty = true;
i = k+1;
return dirty;
}
}
