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A better similarity ranking algorithm for variable length strings

Simon White of Catalysoft wrote an article about a very clever algorithm that compares adjacent character pairs that works really well for my purposes:

http://www.catalysoft.com/articles/StrikeAMatch.html

Simon has a Java version of the algorithm and below I wrote a PL/Ruby version of it (taken from the plain ruby version done in the related forum entry comment by Mark Wong-VanHaren) so that I can use it in my PostgreSQL queries:

CREATE FUNCTION string_similarity(str1 varchar, str2 varchar)
RETURNS float8 AS '

str1.downcase! 
pairs1 = (0..str1.length-2).collect {|i| str1[i,2]}.reject {
  |pair| pair.include? " "}
str2.downcase! 
pairs2 = (0..str2.length-2).collect {|i| str2[i,2]}.reject {
  |pair| pair.include? " "}
union = pairs1.size + pairs2.size 
intersection = 0 
pairs1.each do |p1| 
  0.upto(pairs2.size-1) do |i| 
    if p1 == pairs2[i] 
      intersection += 1 
      pairs2.slice!(i) 
      break 
    end 
  end 
end 
(2.0 * intersection) / union

' LANGUAGE 'plruby';

Works like a charm!


marzagao's answer is great. I converted it to C# so I thought I'd post it here:

Pastebin Link

/// <summary>
/// This class implements string comparison algorithm
/// based on character pair similarity
/// Source: http://www.catalysoft.com/articles/StrikeAMatch.html
/// </summary>
public class SimilarityTool
{
    /// <summary>
    /// Compares the two strings based on letter pair matches
    /// </summary>
    /// <param name="str1"></param>
    /// <param name="str2"></param>
    /// <returns>The percentage match from 0.0 to 1.0 where 1.0 is 100%</returns>
    public double CompareStrings(string str1, string str2)
    {
        List<string> pairs1 = WordLetterPairs(str1.ToUpper());
        List<string> pairs2 = WordLetterPairs(str2.ToUpper());

        int intersection = 0;
        int union = pairs1.Count + pairs2.Count;

        for (int i = 0; i < pairs1.Count; i++)
        {
            for (int j = 0; j < pairs2.Count; j++)
            {
                if (pairs1[i] == pairs2[j])
                {
                    intersection++;
                    pairs2.RemoveAt(j);//Must remove the match to prevent "GGGG" from appearing to match "GG" with 100% success

                    break;
                }
            }
        }

        return (2.0 * intersection) / union;
    }

    /// <summary>
    /// Gets all letter pairs for each
    /// individual word in the string
    /// </summary>
    /// <param name="str"></param>
    /// <returns></returns>
    private List<string> WordLetterPairs(string str)
    {
        List<string> AllPairs = new List<string>();

        // Tokenize the string and put the tokens/words into an array
        string[] Words = Regex.Split(str, @"\s");

        // For each word
        for (int w = 0; w < Words.Length; w++)
        {
            if (!string.IsNullOrEmpty(Words[w]))
            {
                // Find the pairs of characters
                String[] PairsInWord = LetterPairs(Words[w]);

                for (int p = 0; p < PairsInWord.Length; p++)
                {
                    AllPairs.Add(PairsInWord[p]);
                }
            }
        }

        return AllPairs;
    }

    /// <summary>
    /// Generates an array containing every 
    /// two consecutive letters in the input string
    /// </summary>
    /// <param name="str"></param>
    /// <returns></returns>
    private string[] LetterPairs(string str)
    {
        int numPairs = str.Length - 1;

        string[] pairs = new string[numPairs];

        for (int i = 0; i < numPairs; i++)
        {
            pairs[i] = str.Substring(i, 2);
        }

        return pairs;
    }
}

Here is another version of marzagao's answer, this one written in Python:

def get_bigrams(string):
    """
    Take a string and return a list of bigrams.
    """
    s = string.lower()
    return [s[i:i+2] for i in list(range(len(s) - 1))]

def string_similarity(str1, str2):
    """
    Perform bigram comparison between two strings
    and return a percentage match in decimal form.
    """
    pairs1 = get_bigrams(str1)
    pairs2 = get_bigrams(str2)
    union  = len(pairs1) + len(pairs2)
    hit_count = 0
    for x in pairs1:
        for y in pairs2:
            if x == y:
                hit_count += 1
                break
    return (2.0 * hit_count) / union

if __name__ == "__main__":
    """
    Run a test using the example taken from:
    http://www.catalysoft.com/articles/StrikeAMatch.html
    """
    w1 = 'Healed'
    words = ['Heard', 'Healthy', 'Help', 'Herded', 'Sealed', 'Sold']

    for w2 in words:
        print('Healed --- ' + w2)
        print(string_similarity(w1, w2))
        print()

A shorter version of John Rutledge's answer:

def get_bigrams(string):
    '''
    Takes a string and returns a list of bigrams
    '''
    s = string.lower()
    return {s[i:i+2] for i in xrange(len(s) - 1)}

def string_similarity(str1, str2):
    '''
    Perform bigram comparison between two strings
    and return a percentage match in decimal form
    '''
    pairs1 = get_bigrams(str1)
    pairs2 = get_bigrams(str2)
    return (2.0 * len(pairs1 & pairs2)) / (len(pairs1) + len(pairs2))