LevensteinDistance - Commons Lang 3.0 API

问题

With Commons Lang api I can calculate the similarity between two strings through the LevensteinDistance. The result is the number of changes needed to change one string into another. I wish the result was within the range from 0 to 1, where it would be easier to identify the similarity between the strings. The result would be closer to 0 great similarity. Is it possible?

Below the example I'm using:

public class TesteLevenstein {

    public static void main(String[] args) {      

        int distance1 = StringUtils.getLevenshteinDistance("Boat", "Coat");
        int distance2 = StringUtils.getLevenshteinDistance("Remember", "Alamo");
        int distance3 = StringUtils.getLevenshteinDistance("Steve", "Stereo");

        System.out.println("distance(Boat, Coat): " + distance1);
        System.out.println("distance(Remember, Alamo): " + distance2);
        System.out.println("distance(Steve, Stereo): " + distance3);        

    }
}

Thanks!

回答1:

Just divide by some number. The question is what number? Probably the maximum possible distance for the given pair of strings. I think that's the length of the longer string (ie all the characters are different, plus a few more were added, compared with the shorter string).

来源：https://stackoverflow.com/questions/6629712/levensteindistance-commons-lang-3-0-api