N00programmer
N00programmer

Reputation: 1141

Similarity Score - Levenshtein

I implemented the Levenshtein algorithm in Java and am now getting the corrections made by the algorithm, a.k.a. the cost. This does help a little but not much since I want the results as a percentage.

So I want to know how to calculate those similarity points.

I would also like to know how you people do it and why.

Upvotes: 30

Views: 34130

Answers (6)

Alex
Alex

Reputation: 4413

LevenshteinDistance

It can be used through maven dependency

I do think it is better to use this implementation than write your own one.

<dependency>
    <groupId>org.apache.commons</groupId>
    <artifactId>commons-text</artifactId>
    <version>1.3</version>
</dependency>

As an example, have a look at code below

import org.apache.commons.text.similarity.LevenshteinDistance;

public class MetricUtils {
    private static LevenshteinDistance lv = new LevenshteinDistance();

    public static void main(String[] args) {
        String s = "running";
        String s1 = "runninh";
        System.out.println(levensteinRatio(s, s1));
    }

    public static double levensteinRatio(String s, String s1) {
        return 1 - ((double) lv.apply(s, s1)) / Math.max(s.length(), s1.length());
    }
}

Upvotes: 3

userx
userx

Reputation: 876

To calculate score, you need max possible cost(insert+drop+substitute). Then use below formula -

score = 1 - actual_cost/max_possible_cost

See this for reference - Levenshtein Score Calculation Func

Upvotes: -1

Ralph
Ralph

Reputation: 120781

The Levenshtein distance between two strings is defined as the minimum number of edits needed to transform one string into the other, with the allowable edit operations being insertion, deletion, or substitution of a single character. (Wikipedia)

  • So a Levenshtein distance of 0 means: both strings are equal
  • The maximum Levenshtein distance (all chars are different) is max(string1.length, string2.length)

So if you need a percentage, you have to use this to points to scale. For example:

"Hallo", "Hello" -> Levenstein distance 1 Max Levenstein distance for this two strings is: 5. So the 20% of the characters do not match.

String s1 = "Hallo";
String s2 = "Hello";
int lfd = calculateLevensteinDistance(s1, s2);
double ratio = ((double) lfd) / (Math.max(s1.length, s2.length));

Upvotes: 43

Vishal Tathe
Vishal Tathe

Reputation: 190

 // Refer This: 100% working

public class demo 
{
public static void main(String[] args) 
{
    String str1, str2;

    str1="12345";
    str2="122345";


    int re=pecentageOfTextMatch(str1, str2);
    System.out.println("Matching Percent"+re);
}

public static int pecentageOfTextMatch(String s0, String s1) 
{                       // Trim and remove duplicate spaces
    int percentage = 0;
    s0 = s0.trim().replaceAll("\\s+", " ");
    s1 = s1.trim().replaceAll("\\s+", " ");
    percentage=(int) (100 - (float) LevenshteinDistance(s0, s1) * 100 / (float) (s0.length() + s1.length()));
    return percentage;
}

public static int LevenshteinDistance(String s0, String s1) {

    int len0 = s0.length() + 1;
    int len1 = s1.length() + 1;  
    // the array of distances
    int[] cost = new int[len0];
    int[] newcost = new int[len0];

    // initial cost of skipping prefix in String s0
    for (int i = 0; i < len0; i++)
        cost[i] = i;

    // dynamically computing the array of distances

    // transformation cost for each letter in s1
    for (int j = 1; j < len1; j++) {

        // initial cost of skipping prefix in String s1
        newcost[0] = j - 1;

        // transformation cost for each letter in s0
        for (int i = 1; i < len0; i++) {

            // matching current letters in both strings
            int match = (s0.charAt(i - 1) == s1.charAt(j - 1)) ? 0 : 1;

            // computing cost for each transformation
            int cost_replace = cost[i - 1] + match;
            int cost_insert = cost[i] + 1;
            int cost_delete = newcost[i - 1] + 1;

            // keep minimum cost
            newcost[i] = Math.min(Math.min(cost_insert, cost_delete),
                    cost_replace);
        }

        // swap cost/newcost arrays
        int[] swap = cost;
        cost = newcost;
        newcost = swap;
    }

    // the distance is the cost for transforming all letters in both strings
    return cost[len0 - 1];
}

}

Upvotes: 4

Roman
Roman

Reputation: 66156

You can download Apache Commons StringUtils and investigate (and maybe use) their implementation of Levenshtein distance algorithm.

Upvotes: 19

Donal Fellows
Donal Fellows

Reputation: 137567

The maximum value of the Levenshtein difference between two strings would be the maximum of the length of the two strings. (That corresponds to a change of symbol for each of the characters up to the length of the shorter string, plus inserts or deletes depending on whether you're going from shorter to longer or vice versa.) Given that, the similarity of the two strings must be the ratio between that maximum and the difference between that maximum and the actual Levenshtein difference.

Implementations of the Levenshtein algorithm tend to not record what those edits should be, but it shouldn't be that hard to calculate given the abstract algorithm on the Wikipedia page.

Upvotes: 0

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