Optimizing Jaro-Winkler algorithm
Posted
by Pentium10
on Stack Overflow
See other posts from Stack Overflow
or by Pentium10
Published on 2010-05-17T12:01:15Z
Indexed on
2010/05/17
12:40 UTC
Read the original article
Hit count: 808
I have this code for Jaro-Winkler algorithm taken from this website. I need to run 150,000 times to get distance between differences. It takes a long time, as I run on an Android mobile device.
Can it be optimized more?
public class Jaro {
/**
* gets the similarity of the two strings using Jaro distance.
*
* @param string1 the first input string
* @param string2 the second input string
* @return a value between 0-1 of the similarity
*/
public float getSimilarity(final String string1, final String string2) {
//get half the length of the string rounded up - (this is the distance used for acceptable transpositions)
final int halflen = ((Math.min(string1.length(), string2.length())) / 2) + ((Math.min(string1.length(), string2.length())) % 2);
//get common characters
final StringBuffer common1 = getCommonCharacters(string1, string2, halflen);
final StringBuffer common2 = getCommonCharacters(string2, string1, halflen);
//check for zero in common
if (common1.length() == 0 || common2.length() == 0) {
return 0.0f;
}
//check for same length common strings returning 0.0f is not the same
if (common1.length() != common2.length()) {
return 0.0f;
}
//get the number of transpositions
int transpositions = 0;
int n=common1.length();
for (int i = 0; i < n; i++) {
if (common1.charAt(i) != common2.charAt(i))
transpositions++;
}
transpositions /= 2.0f;
//calculate jaro metric
return (common1.length() / ((float) string1.length()) +
common2.length() / ((float) string2.length()) +
(common1.length() - transpositions) / ((float) common1.length())) / 3.0f;
}
/**
* returns a string buffer of characters from string1 within string2 if they are of a given
* distance seperation from the position in string1.
*
* @param string1
* @param string2
* @param distanceSep
* @return a string buffer of characters from string1 within string2 if they are of a given
* distance seperation from the position in string1
*/
private static StringBuffer getCommonCharacters(final String string1, final String string2, final int distanceSep) {
//create a return buffer of characters
final StringBuffer returnCommons = new StringBuffer();
//create a copy of string2 for processing
final StringBuffer copy = new StringBuffer(string2);
//iterate over string1
int n=string1.length();
int m=string2.length();
for (int i = 0; i < n; i++) {
final char ch = string1.charAt(i);
//set boolean for quick loop exit if found
boolean foundIt = false;
//compare char with range of characters to either side
for (int j = Math.max(0, i - distanceSep); !foundIt && j < Math.min(i + distanceSep, m - 1); j++) {
//check if found
if (copy.charAt(j) == ch) {
foundIt = true;
//append character found
returnCommons.append(ch);
//alter copied string2 for processing
copy.setCharAt(j, (char)0);
}
}
}
return returnCommons;
}
}
I mention that in the whole process I make just instance of the script, so only once
jaro= new Jaro();
If you are going to test and need examples so not break the script, you will find it here, in another thread for python optimization.
© Stack Overflow or respective owner