Compare the similarity of 2 given addresses

Hi @Jonathan ,

The Levenshtein Distance implementation you shared is a valid approach and works well for finding small spelling differences, typos, or character-level changes between two strings.

However, for address comparison it has some limitations. Since it compares the entire address character by character, it can produce a lower similarity score when the same address components appear in a different order

To better align with the requirement that approximately 80% of the address words should match regardless of their position, I implemented a hybrid approach:

public static double WordSimilarity(string first, string second)
        {
            MatchCollection firstWords = Regex.Matches(first ?? "", @"[\p{L}\p{N}]+");
            MatchCollection secondWords = Regex.Matches(second ?? "", @"[\p{L}\p{N}]+");
 
            if (firstWords.Count == 0 || secondWords.Count == 0)
                return 0;
 
            var firstMatched = new bool[firstWords.Count];
            var secondMatched = new bool[secondWords.Count];
            int matches = 0;
 
            for (int secondIndex = 0; secondIndex < secondWords.Count; secondIndex++)
            {
                for (int firstIndex = 0; firstIndex < firstWords.Count; firstIndex++)
                {
                    if (!firstMatched[firstIndex] &&
                        string.Equals(firstWords[firstIndex].Value,
                            secondWords[secondIndex].Value,
                            StringComparison.OrdinalIgnoreCase))
                    {
                        firstMatched[firstIndex] = true;
                        secondMatched[secondIndex] = true;
                        matches++;
                        break;
                    }
                }
            }
 
            for (int secondIndex = 0; secondIndex < secondWords.Count; secondIndex++)
            {
                if (secondMatched[secondIndex])
                    continue;
 
                string secondWord = secondWords[secondIndex].Value;
                if (secondWord.Length < 4 || ContainsDigit(secondWord))
                    continue;
 
                for (int firstIndex = 0; firstIndex < firstWords.Count; firstIndex++)
                {
                    string firstWord = firstWords[firstIndex].Value;
                    if (!firstMatched[firstIndex] && firstWord.Length >= 4 &&
                        !ContainsDigit(firstWord) && WithinOneEdit(firstWord, secondWord))
                    {
                        firstMatched[firstIndex] = true;
                        matches++;
                        break;
                    }
                }
            }
 
            return (double)matches / Math.Max(firstWords.Count, secondWords.Count);
        }
 
        public static bool IsMatch(string first, string second)
        {
            return WordSimilarity(first, second) >= 0.8;
        }
 
        private static bool ContainsDigit(string word)
        {
            foreach (char character in word)
            {
                if (char.IsDigit(character))
                    return true;
            }
            return false;
        }
 
        private static bool WithinOneEdit(string first, string second)
        {
            if (Math.Abs(first.Length - second.Length) > 1)
                return false;
 
            int firstIndex = 0;
            int secondIndex = 0;
            int edits = 0;
 
            while (firstIndex < first.Length && secondIndex < second.Length)
            {
                if (char.ToUpperInvariant(first[firstIndex]) == char.ToUpperInvariant(second[secondIndex]))
                {
                    firstIndex++;
                    secondIndex++;
                    continue;
                }
 
                if (++edits > 1)
                    return false;
 
                if (first.Length >= second.Length)
                    firstIndex++;
                if (second.Length >= first.Length)
                    secondIndex++;
            }
 
            return edits + (first.Length - firstIndex) + (second.Length - secondIndex) <= 1;
        }
    }

Here's what the implementation does:

1. Split both addresses into individual words and numbers -The regular expression extracts each address component separately instead of treating the address as one long string.
2. Perform exact word matching first -The code compares each word from one address against the other. -Matching is case-insensitive. -The position of the words does not matter, so addresses containing the same components in different orders can still achieve a high score.
3. Perform a second pass for unmatched words -If a word was not matched exactly, the code checks whether the words differ by only a single edit using the WithinOneEdit() method.
4. Avoid fuzzy matching on short or numeric values

Compared to applying Levenshtein Distance to the entire address string, I believe this is more suitable for address matching scenarios.

If you found my response helpful or informative, I would greatly appreciate it if you could follow this guide for your confirmation.

Thank you.

Hi @Jonathan

Please try the below method will achieve the your result.

public static double WordMatchPercentage(string a, string b)
{
 var words1 = a.ToLower().Split(' ', ',', '.')
 .Where(x => !string.IsNullOrWhiteSpace(x))
 .ToHashSet();
 
 var words2 = b.ToLower().Split(' ', ',', '.')
 .Where(x => !string.IsNullOrWhiteSpace(x))
 .ToHashSet();
 
 int matched = words1.Intersect(words2).Count();
 
 int maxWords = Math.Max(words1.Count, words2.Count);
 
 return (double)matched / maxWords * 100;
}

Thanks.