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info.debatty.java.stringsimilarity.RatcliffObershelp Maven / Gradle / Ivy
/*
* The MIT License
*
* Copyright 2015 Thibault Debatty.
*
* Permission is hereby granted, free of charge, to any person obtaining a copy
* of this software and associated documentation files (the "Software"), to deal
* in the Software without restriction, including without limitation the rights
* to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
* copies of the Software, and to permit persons to whom the Software is
* furnished to do so, subject to the following conditions:
*
* The above copyright notice and this permission notice shall be included in
* all copies or substantial portions of the Software.
*
* THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
* IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
* FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
* AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
* LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
* OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
* THE SOFTWARE.
*/
package info.debatty.java.stringsimilarity;
import info.debatty.java.stringsimilarity.interfaces.NormalizedStringSimilarity;
import info.debatty.java.stringsimilarity.interfaces.NormalizedStringDistance;
import java.util.List;
import java.util.ArrayList;
import net.jcip.annotations.Immutable;
/**
* Ratcliff/Obershelp pattern recognition
* The Ratcliff/Obershelp algorithm computes the similarity of two strings a
* the doubled number of matching characters divided by the total number of
* characters in the two strings. Matching characters are those in the longest
* common subsequence plus, recursively, matching characters in the unmatched
* region on either side of the longest common subsequence.
* The Ratcliff/Obershelp distance is computed as 1 - Ratcliff/Obershelp
* similarity.
*
* @author Ligi https://github.com/dxpux (as a patch for fuzzystring)
* Ported to java from .net by denmase
*/
@Immutable
public class RatcliffObershelp implements
NormalizedStringSimilarity, NormalizedStringDistance {
/**
* Compute the Ratcliff-Obershelp similarity between strings.
*
* @param s1 The first string to compare.
* @param s2 The second string to compare.
* @return The RatcliffObershelp similarity in the range [0, 1]
* @throws NullPointerException if s1 or s2 is null.
*/
@Override
public final double similarity(final String s1, final String s2) {
if (s1 == null) {
throw new NullPointerException("s1 must not be null");
}
if (s2 == null) {
throw new NullPointerException("s2 must not be null");
}
if (s1.equals(s2)) {
return 1.0d;
}
List matches = getMatchList(s1, s2);
int sum_of_matches = 0;
for (String match : matches) {
sum_of_matches += match.length();
}
return 2.0d * sum_of_matches / (s1.length() + s2.length());
}
/**
* Return 1 - similarity.
*
* @param s1 The first string to compare.
* @param s2 The second string to compare.
* @return 1 - similarity
* @throws NullPointerException if s1 or s2 is null.
*/
@Override
public final double distance(final String s1, final String s2) {
return 1.0d - similarity(s1, s2);
}
private static List getMatchList(final String s1, final String s2) {
List list = new ArrayList();
String match = frontMaxMatch(s1, s2);
if (match.length() > 0) {
String frontsource = s1.substring(0, s1.indexOf(match));
String fronttarget = s2.substring(0, s2.indexOf(match));
List frontqueue = getMatchList(frontsource, fronttarget);
String endsource = s1.substring(s1.indexOf(match) + match.length());
String endtarget = s2.substring(s2.indexOf(match) + match.length());
List endqueue = getMatchList(endsource, endtarget);
list.add(match);
list.addAll(frontqueue);
list.addAll(endqueue);
}
return list;
}
private static String frontMaxMatch(final String s1, final String s2) {
int longest = 0;
String longestsubstring = "";
for (int i = 0; i < s1.length(); ++i) {
for (int j = i + 1; j <= s1.length(); ++j) {
String substring = s1.substring(i, j);
if (s2.contains(substring) && substring.length() > longest) {
longest = substring.length();
longestsubstring = substring;
}
}
}
return longestsubstring;
}
}
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