A better similarity ranking algorithm for variable length strings - string-matching

Question

I'm looking for a string similarity algorithm that yields better results on variable length strings than the ones that are usually suggested (levenshtein distance, soundex, etc).
For example,
Given string A: "Robert",
Then string B: "Amy Robertson"
would be a better match than
String C: "Richard"
Also, preferably, this algorithm should be language agnostic (also works in languages other than English).

Answer 1

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!

Answer 2

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;
}
}

Answer 3

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()

Answer 4

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))

Answer 5

Here's my PHP implementation of suggested StrikeAMatch algorithm, by Simon White. the advantages (like it says in the link) are:
A true reflection of lexical similarity - strings with small differences should be recognised as being similar. In particular, a significant substring overlap should point to a high level of similarity between the strings.
A robustness to changes of word order - two strings which contain the same words, but in a different order, should be recognised as being similar. On the other hand, if one string is just a random anagram of the characters contained in the other, then it should (usually) be recognised as dissimilar.
Language Independence - the algorithm should work not only in English, but in many different languages.
<?php
/**
* LetterPairSimilarity algorithm implementation in PHP
* #author Igal Alkon
* #link http://www.catalysoft.com/articles/StrikeAMatch.html
*/
class LetterPairSimilarity
{
/**
* #param $str
* #return mixed
*/
private function wordLetterPairs($str)
{
$allPairs = array();
// Tokenize the string and put the tokens/words into an array
$words = explode(' ', $str);
// For each word
for ($w = 0; $w < count($words); $w++)
{
// Find the pairs of characters
$pairsInWord = $this->letterPairs($words[$w]);
for ($p = 0; $p < count($pairsInWord); $p++)
{
$allPairs[] = $pairsInWord[$p];
}
}
return $allPairs;
}
/**
* #param $str
* #return array
*/
private function letterPairs($str)
{
$numPairs = mb_strlen($str)-1;
$pairs = array();
for ($i = 0; $i < $numPairs; $i++)
{
$pairs[$i] = mb_substr($str,$i,2);
}
return $pairs;
}
/**
* #param $str1
* #param $str2
* #return float
*/
public function compareStrings($str1, $str2)
{
$pairs1 = $this->wordLetterPairs(strtoupper($str1));
$pairs2 = $this->wordLetterPairs(strtoupper($str2));
$intersection = 0;
$union = count($pairs1) + count($pairs2);
for ($i=0; $i < count($pairs1); $i++)
{
$pair1 = $pairs1[$i];
$pairs2 = array_values($pairs2);
for($j = 0; $j < count($pairs2); $j++)
{
$pair2 = $pairs2[$j];
if ($pair1 === $pair2)
{
$intersection++;
unset($pairs2[$j]);
break;
}
}
}
return (2.0*$intersection)/$union;
}
}

Answer 6

This discussion has been really helpful, thanks. I converted the algorithm to VBA for use with Excel and wrote a few versions of a worksheet function, one for simple comparison of a pair of strings, the other for comparing one string to a range/array of strings. The strSimLookup version returns either the last best match as a string, array index, or similarity metric.
This implementation produces the same results listed in the Amazon example on Simon White's website with a few minor exceptions on low-scoring matches; not sure where the difference creeps in, could be VBA's Split function, but I haven't investigated as it's working fine for my purposes.
'Implements functions to rate how similar two strings are on
'a scale of 0.0 (completely dissimilar) to 1.0 (exactly similar)
'Source:  http://www.catalysoft.com/articles/StrikeAMatch.html
'Author: Bob Chatham, bob.chatham at gmail.com
'9/12/2010
Option Explicit
Public Function stringSimilarity(str1 As String, str2 As String) As Variant
'Simple version of the algorithm that computes the similiarity metric
'between two strings.
'NOTE: This verision is not efficient to use if you're comparing one string
'with a range of other values as it will needlessly calculate the pairs for the
'first string over an over again; use the array-optimized version for this case.
Dim sPairs1 As Collection
Dim sPairs2 As Collection
Set sPairs1 = New Collection
Set sPairs2 = New Collection
WordLetterPairs str1, sPairs1
WordLetterPairs str2, sPairs2
stringSimilarity = SimilarityMetric(sPairs1, sPairs2)
Set sPairs1 = Nothing
Set sPairs2 = Nothing
End Function
Public Function strSimA(str1 As Variant, rRng As Range) As Variant
'Return an array of string similarity indexes for str1 vs every string in input range rRng
Dim sPairs1 As Collection
Dim sPairs2 As Collection
Dim arrOut As Variant
Dim l As Long, j As Long
Set sPairs1 = New Collection
WordLetterPairs CStr(str1), sPairs1
l = rRng.Count
ReDim arrOut(1 To l)
For j = 1 To l
Set sPairs2 = New Collection
WordLetterPairs CStr(rRng(j)), sPairs2
arrOut(j) = SimilarityMetric(sPairs1, sPairs2)
Set sPairs2 = Nothing
Next j
strSimA = Application.Transpose(arrOut)
End Function
Public Function strSimLookup(str1 As Variant, rRng As Range, Optional returnType) As Variant
'Return either the best match or the index of the best match
'depending on returnTYype parameter) between str1 and strings in rRng)
' returnType = 0 or omitted: returns the best matching string
' returnType = 1 : returns the index of the best matching string
' returnType = 2 : returns the similarity metric
Dim sPairs1 As Collection
Dim sPairs2 As Collection
Dim metric, bestMetric As Double
Dim i, iBest As Long
Const RETURN_STRING As Integer = 0
Const RETURN_INDEX As Integer = 1
Const RETURN_METRIC As Integer = 2
If IsMissing(returnType) Then returnType = RETURN_STRING
Set sPairs1 = New Collection
WordLetterPairs CStr(str1), sPairs1
bestMetric = -1
iBest = -1
For i = 1 To rRng.Count
Set sPairs2 = New Collection
WordLetterPairs CStr(rRng(i)), sPairs2
metric = SimilarityMetric(sPairs1, sPairs2)
If metric > bestMetric Then
bestMetric = metric
iBest = i
End If
Set sPairs2 = Nothing
Next i
If iBest = -1 Then
strSimLookup = CVErr(xlErrValue)
Exit Function
End If
Select Case returnType
Case RETURN_STRING
strSimLookup = CStr(rRng(iBest))
Case RETURN_INDEX
strSimLookup = iBest
Case Else
strSimLookup = bestMetric
End Select
End Function
Public Function strSim(str1 As String, str2 As String) As Variant
Dim ilen, iLen1, ilen2 As Integer
iLen1 = Len(str1)
ilen2 = Len(str2)
If iLen1 >= ilen2 Then ilen = ilen2 Else ilen = iLen1
strSim = stringSimilarity(Left(str1, ilen), Left(str2, ilen))
End Function
Sub WordLetterPairs(str As String, pairColl As Collection)
'Tokenize str into words, then add all letter pairs to pairColl
Dim Words() As String
Dim word, nPairs, pair As Integer
Words = Split(str)
If UBound(Words) < 0 Then
Set pairColl = Nothing
Exit Sub
End If
For word = 0 To UBound(Words)
nPairs = Len(Words(word)) - 1
If nPairs > 0 Then
For pair = 1 To nPairs
pairColl.Add Mid(Words(word), pair, 2)
Next pair
End If
Next word
End Sub
Private Function SimilarityMetric(sPairs1 As Collection, sPairs2 As Collection) As Variant
'Helper function to calculate similarity metric given two collections of letter pairs.
'This function is designed to allow the pair collections to be set up separately as needed.
'NOTE: sPairs2 collection will be altered as pairs are removed; copy the collection
'if this is not the desired behavior.
'Also assumes that collections will be deallocated somewhere else
Dim Intersect As Double
Dim Union As Double
Dim i, j As Long
If sPairs1.Count = 0 Or sPairs2.Count = 0 Then
SimilarityMetric = CVErr(xlErrNA)
Exit Function
End If
Union = sPairs1.Count + sPairs2.Count
Intersect = 0
For i = 1 To sPairs1.Count
For j = 1 To sPairs2.Count
If StrComp(sPairs1(i), sPairs2(j)) = 0 Then
Intersect = Intersect + 1
sPairs2.Remove j
Exit For
End If
Next j
Next i
SimilarityMetric = (2 * Intersect) / Union
End Function

Answer 7

I'm sorry, the answer was not invented by the author. This is a well known algorithm that was first present by Digital Equipment Corporation and is often referred to as shingling.
http://www.hpl.hp.com/techreports/Compaq-DEC/SRC-TN-1997-015.pdf

Answer 8

I translated Simon White's algorithm to PL/pgSQL. This is my contribution.
<!-- language: lang-sql -->
create or replace function spt1.letterpairs(in p_str varchar)
returns varchar as
$$
declare
v_numpairs integer := length(p_str)-1;
v_pairs varchar[];
begin
for i in 1 .. v_numpairs loop
v_pairs[i] := substr(p_str, i, 2);
end loop;
return v_pairs;
end;
$$ language 'plpgsql';
--===================================================================
create or replace function spt1.wordletterpairs(in p_str varchar)
returns varchar as
$$
declare
v_allpairs varchar[];
v_words varchar[];
v_pairsinword varchar[];
begin
v_words := regexp_split_to_array(p_str, '[[:space:]]');
for i in 1 .. array_length(v_words, 1) loop
v_pairsinword := spt1.letterpairs(v_words[i]);
if v_pairsinword is not null then
for j in 1 .. array_length(v_pairsinword, 1) loop
v_allpairs := v_allpairs || v_pairsinword[j];
end loop;
end if;
end loop;
return v_allpairs;
end;
$$ language 'plpgsql';
--===================================================================
create or replace function spt1.arrayintersect(ANYARRAY, ANYARRAY)
returns anyarray as
$$
select array(select unnest($1) intersect select unnest($2))
$$ language 'sql';
--===================================================================
create or replace function spt1.comparestrings(in p_str1 varchar, in p_str2 varchar)
returns float as
$$
declare
v_pairs1 varchar[];
v_pairs2 varchar[];
v_intersection integer;
v_union integer;
begin
v_pairs1 := wordletterpairs(upper(p_str1));
v_pairs2 := wordletterpairs(upper(p_str2));
v_union := array_length(v_pairs1, 1) + array_length(v_pairs2, 1);
v_intersection := array_length(arrayintersect(v_pairs1, v_pairs2), 1);
return (2.0 * v_intersection / v_union);
end;
$$ language 'plpgsql';

Answer 9

A version in beautiful Scala:
def pairDistance(s1: String, s2: String): Double = {
def strToPairs(s: String, acc: List[String]): List[String] = {
if (s.size < 2) acc
else strToPairs(s.drop(1),
if (s.take(2).contains(" ")) acc else acc ::: List(s.take(2)))
}
val lst1 = strToPairs(s1.toUpperCase, List())
val lst2 = strToPairs(s2.toUpperCase, List())
(2.0 * lst2.intersect(lst1).size) / (lst1.size + lst2.size)
}

Answer 10

String Similarity Metrics contains an overview of many different metrics used in string comparison (Wikipedia has an overview as well). Much of these metrics is implemented in a library simmetrics.
Yet another example of metric, not included in the given overview is for example compression distance (attempting to approximate the Kolmogorov's complexity), which can be used for a bit longer texts than the one you presented.
You might also consider looking at a much broader subject of Natural Language Processing. These R packages can get you started quickly (or at least give some ideas).
And one last edit - search the other questions on this subject at SO, there are quite a few related ones.

Answer 11

A faster PHP version of the algorithm:
/**
*
* #param $str
* #return mixed
*/
private static function wordLetterPairs ($str)
{
$allPairs = array();
// Tokenize the string and put the tokens/words into an array
$words = explode(' ', $str);
// For each word
for ($w = 0; $w < count($words); $w ++) {
// Find the pairs of characters
$pairsInWord = self::letterPairs($words[$w]);
for ($p = 0; $p < count($pairsInWord); $p ++) {
$allPairs[$pairsInWord[$p]] = $pairsInWord[$p];
}
}
return array_values($allPairs);
}
/**
*
* #param $str
* #return array
*/
private static function letterPairs ($str)
{
$numPairs = mb_strlen($str) - 1;
$pairs = array();
for ($i = 0; $i < $numPairs; $i ++) {
$pairs[$i] = mb_substr($str, $i, 2);
}
return $pairs;
}
/**
*
* #param $str1
* #param $str2
* #return float
*/
public static function compareStrings ($str1, $str2)
{
$pairs1 = self::wordLetterPairs(mb_strtolower($str1));
$pairs2 = self::wordLetterPairs(mb_strtolower($str2));
$union = count($pairs1) + count($pairs2);
$intersection = count(array_intersect($pairs1, $pairs2));
return (2.0 * $intersection) / $union;
}
For the data I had (approx 2300 comparisons) I had a running time of 0.58sec with Igal Alkon solution versus 0.35sec with mine.

Answer 12

Posting marzagao's answer in C99, inspired by these algorithms
double dice_match(const char *string1, const char *string2) {
//check fast cases
if (((string1 != NULL) && (string1[0] == '\0')) ||
((string2 != NULL) && (string2[0] == '\0'))) {
return 0;
}
if (string1 == string2) {
return 1;
}
size_t strlen1 = strlen(string1);
size_t strlen2 = strlen(string2);
if (strlen1 < 2 || strlen2 < 2) {
return 0;
}
size_t length1 = strlen1 - 1;
size_t length2 = strlen2 - 1;
double matches = 0;
int i = 0, j = 0;
//get bigrams and compare
while (i < length1 && j < length2) {
char a[3] = {string1[i], string1[i + 1], '\0'};
char b[3] = {string2[j], string2[j + 1], '\0'};
int cmp = strcmpi(a, b);
if (cmp == 0) {
matches += 2;
}
i++;
j++;
}
return matches / (length1 + length2);
}
Some tests based on the original article:
#include <stdio.h>
void article_test1() {
char *string1 = "FRANCE";
char *string2 = "FRENCH";
printf("====%s====\n", __func__);
printf("%2.f%% == 40%%\n", dice_match(string1, string2) * 100);
}
void article_test2() {
printf("====%s====\n", __func__);
char *string = "Healed";
char *ss[] = {"Heard", "Healthy", "Help",
"Herded", "Sealed", "Sold"};
int correct[] = {44, 55, 25, 40, 80, 0};
for (int i = 0; i < 6; ++i) {
printf("%2.f%% == %d%%\n", dice_match(string, ss[i]) * 100, correct[i]);
}
}
void multicase_test() {
char *string1 = "FRaNcE";
char *string2 = "fREnCh";
printf("====%s====\n", __func__);
printf("%2.f%% == 40%%\n", dice_match(string1, string2) * 100);
}
void gg_test() {
char *string1 = "GG";
char *string2 = "GGGGG";
printf("====%s====\n", __func__);
printf("%2.f%% != 100%%\n", dice_match(string1, string2) * 100);
}
int main() {
article_test1();
article_test2();
multicase_test();
gg_test();
return 0;
}

Answer 13

Here is the R version:
get_bigrams <- function(str)
{
lstr = tolower(str)
bigramlst = list()
for(i in 1:(nchar(str)-1))
{
bigramlst[[i]] = substr(str, i, i+1)
}
return(bigramlst)
}
str_similarity <- function(str1, str2)
{
pairs1 = get_bigrams(str1)
pairs2 = get_bigrams(str2)
unionlen = length(pairs1) + length(pairs2)
hit_count = 0
for(x in 1:length(pairs1)){
for(y in 1:length(pairs2)){
if (pairs1[[x]] == pairs2[[y]])
hit_count = hit_count + 1
}
}
return ((2.0 * hit_count) / unionlen)
}

Answer 14

Building on Michael La Voie's awesome C# version, as per the request to make it an extension method, here is what I came up with. The primary benefit of doing it this way is that you can sort a Generic List by the percent match. For example, consider you have a string field named "City" in your object. A user searches for "Chester" and you want to return results in descending order of match. For example, you want literal matches of Chester to show up before Rochester. To do this, add two new properties to your object:
public string SearchText { get; set; }
public double PercentMatch
{
get
{
return City.ToUpper().PercentMatchTo(this.SearchText.ToUpper());
}
}
Then on each object, set the SearchText to what the user searched for. Then you can sort it easily with something like:
zipcodes = zipcodes.OrderByDescending(x => x.PercentMatch);
Here's the slight modification to make it an extension method:
/// <summary>
/// This class implements string comparison algorithm
/// based on character pair similarity
/// Source: http://www.catalysoft.com/articles/StrikeAMatch.html
/// </summary>
public static double PercentMatchTo(this 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 static 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 static 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;
}

Answer 15

My JavaScript implementation takes a string or array of strings, and an optional floor (the default floor is 0.5). If you pass it a string, it will return true or false depending on whether or not the string's similarity score is greater than or equal to the floor. If you pass it an array of strings, it will return an array of those strings whose similarity score is greater than or equal to the floor, sorted by score.
Examples:
'Healed'.fuzzy('Sealed'); // returns true
'Healed'.fuzzy('Help'); // returns false
'Healed'.fuzzy('Help', 0.25); // returns true
'Healed'.fuzzy(['Sold', 'Herded', 'Heard', 'Help', 'Sealed', 'Healthy']);
// returns ["Sealed", "Healthy"]
'Healed'.fuzzy(['Sold', 'Herded', 'Heard', 'Help', 'Sealed', 'Healthy'], 0);
// returns ["Sealed", "Healthy", "Heard", "Herded", "Help", "Sold"]
Here it is:
(function(){
var default_floor = 0.5;
function pairs(str){
var pairs = []
, length = str.length - 1
, pair;
str = str.toLowerCase();
for(var i = 0; i < length; i++){
pair = str.substr(i, 2);
if(!/\s/.test(pair)){
pairs.push(pair);
}
}
return pairs;
}
function similarity(pairs1, pairs2){
var union = pairs1.length + pairs2.length
, hits = 0;
for(var i = 0; i < pairs1.length; i++){
for(var j = 0; j < pairs2.length; j++){
if(pairs1[i] == pairs2[j]){
pairs2.splice(j--, 1);
hits++;
break;
}
}
}
return 2*hits/union || 0;
}
String.prototype.fuzzy = function(strings, floor){
var str1 = this
, pairs1 = pairs(this);
floor = typeof floor == 'number' ? floor : default_floor;
if(typeof(strings) == 'string'){
return str1.length > 1 && strings.length > 1 && similarity(pairs1, pairs(strings)) >= floor || str1.toLowerCase() == strings.toLowerCase();
}else if(strings instanceof Array){
var scores = {};
strings.map(function(str2){
scores[str2] = str1.length > 1 ? similarity(pairs1, pairs(str2)) : 1*(str1.toLowerCase() == str2.toLowerCase());
});
return strings.filter(function(str){
return scores[str] >= floor;
}).sort(function(a, b){
return scores[b] - scores[a];
});
}
};
})();

Answer 16

The Dice coefficient algorithm (Simon White / marzagao's answer) is implemented in Ruby in the
pair_distance_similar method in the amatch gem
https://github.com/flori/amatch
This gem also contains implementations of a number of approximate matching and string comparison algorithms: Levenshtein edit distance, Sellers edit distance, the Hamming distance, the longest common subsequence length, the longest common substring length, the pair distance metric, the Jaro-Winkler metric.

Answer 17

A Haskell version—feel free to suggest edits because I haven't done much Haskell.
import Data.Char
import Data.List
-- Convert a string into words, then get the pairs of words from that phrase
wordLetterPairs :: String -> [String]
wordLetterPairs s1 = concat $ map pairs $ words s1
-- Converts a String into a list of letter pairs.
pairs :: String -> [String]
pairs [] = []
pairs (x:[]) = []
pairs (x:ys) = [x, head ys]:(pairs ys)
-- Calculates the match rating for two strings
matchRating :: String -> String -> Double
matchRating s1 s2 = (numberOfMatches * 2) / totalLength
where pairsS1 = wordLetterPairs $ map toLower s1
pairsS2 = wordLetterPairs $ map toLower s2
numberOfMatches = fromIntegral $ length $ pairsS1 `intersect` pairsS2
totalLength = fromIntegral $ length pairsS1 + length pairsS2

Answer 18

Clojure:
(require '[clojure.set :refer [intersection]])
(defn bigrams [s]
(->> (split s #"\s+")
(mapcat #(partition 2 1 %))
(set)))
(defn string-similarity [a b]
(let [a-pairs (bigrams a)
b-pairs (bigrams b)
total-count (+ (count a-pairs) (count b-pairs))
match-count (count (intersection a-pairs b-pairs))
similarity (/ (* 2 match-count) total-count)]
similarity))

Answer 19

Here is another version of Similarity based in Sørensen–Dice index (marzagao's answer), this one written in C++11:
/*
* Similarity based in Sørensen–Dice index.
*
* Returns the Similarity between _str1 and _str2.
*/
double similarity_sorensen_dice(const std::string& _str1, const std::string& _str2) {
// Base case: if some string is empty.
if (_str1.empty() || _str2.empty()) {
return 1.0;
}
auto str1 = upper_string(_str1);
auto str2 = upper_string(_str2);
// Base case: if the strings are equals.
if (str1 == str2) {
return 0.0;
}
// Base case: if some string does not have bigrams.
if (str1.size() < 2 || str2.size() < 2) {
return 1.0;
}
// Extract bigrams from str1
auto num_pairs1 = str1.size() - 1;
std::unordered_set<std::string> str1_bigrams;
str1_bigrams.reserve(num_pairs1);
for (unsigned i = 0; i < num_pairs1; ++i) {
str1_bigrams.insert(str1.substr(i, 2));
}
// Extract bigrams from str2
auto num_pairs2 = str2.size() - 1;
std::unordered_set<std::string> str2_bigrams;
str2_bigrams.reserve(num_pairs2);
for (unsigned int i = 0; i < num_pairs2; ++i) {
str2_bigrams.insert(str2.substr(i, 2));
}
// Find the intersection between the two sets.
int intersection = 0;
if (str1_bigrams.size() < str2_bigrams.size()) {
const auto it_e = str2_bigrams.end();
for (const auto& bigram : str1_bigrams) {
intersection += str2_bigrams.find(bigram) != it_e;
}
} else {
const auto it_e = str1_bigrams.end();
for (const auto& bigram : str2_bigrams) {
intersection += str1_bigrams.find(bigram) != it_e;
}
}
// Returns similarity coefficient.
return (2.0 * intersection) / (num_pairs1 + num_pairs2);
}

Answer 20

Why not for a JavaScript implementation, I also explained the algorithm.
Algorithm
Input : France and French.
Map them both to their upper case characters (making the algorithm insensitive to case differences), then split them up into their character pairs:
FRANCE: {FR, RA, AN, NC, CE}
FRENCH: {FR, RE, EN, NC, CH}
Find there intersection:
Result:
Implementation
function similarity(s1, s2) {
const
set1 = pairs(s1.toUpperCase()), // [ FR, RA, AN, NC, CE ]
set2 = pairs(s2.toUpperCase()), // [ FR, RE, EN, NC, CH ]
intersection = set1.filter(x => set2.includes(x)); // [ FR, NC ]
// Tips: Instead of `2` multiply by `200`, To get percentage.
return (intersection.length * 2) / (set1.length + set2.length);
}
function pairs(input) {
const tokenized = [];
for (let i = 0; i < input.length - 1; i++)
tokenized.push(input.substring(i, 2 + i));
return tokenized;
}
console.log(similarity("FRANCE", "FRENCH"));
Ranking Results By ( Word - Similarity )
Sealed - 80%
Healthy - 55%
Heard - 44%
Herded - 40%
Help - 25%
Sold - 0%
From same original source.

Answer 21

What about Levenshtein distance, divided by the length of the first string (or alternatively divided my min/max/avg length of both strings)? That has worked for me so far.

Answer 22

Hey guys i gave this a try in javascript, but I'm new to it, anyone know faster ways to do it?
function get_bigrams(string) {
// Takes a string and returns a list of bigrams
var s = string.toLowerCase();
var v = new Array(s.length-1);
for (i = 0; i< v.length; i++){
v[i] =s.slice(i,i+2);
}
return v;
}
function string_similarity(str1, str2){
/*
Perform bigram comparison between two strings
and return a percentage match in decimal form
*/
var pairs1 = get_bigrams(str1);
var pairs2 = get_bigrams(str2);
var union = pairs1.length + pairs2.length;
var hit_count = 0;
for (x in pairs1){
for (y in pairs2){
if (pairs1[x] == pairs2[y]){
hit_count++;
}
}
}
return ((2.0 * hit_count) / union);
}
var w1 = 'Healed';
var word =['Heard','Healthy','Help','Herded','Sealed','Sold']
for (w2 in word){
console.log('Healed --- ' + word[w2])
console.log(string_similarity(w1,word[w2]));
}

Answer 23

I was looking for pure ruby implementation of the algorithm indicated by #marzagao's answer. Unfortunately, link indicated by #marzagao is broken. In #s01ipsist answer, he indicated ruby gem amatch where implementation is not in pure ruby. So I searchd a little and found gem fuzzy_match which has pure ruby implementation (though this gem use amatch) at here. I hope this will help someone like me.

Answer 24

**I've converted marzagao's answer to Java.**
import org.apache.commons.lang3.StringUtils; //Add a apache commons jar in pom.xml
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
public class SimilarityComparator {
public static void main(String[] args) {
String str0 = "Nischal";
String str1 = "Nischal";
double v = compareStrings(str0, str1);
System.out.println("Similarity betn " + str0 + " and " + str1 + " = " + v);
}
private static double compareStrings(String str1, String str2) {
List<String> pairs1 = wordLetterPairs(str1.toUpperCase());
List<String> pairs2 = wordLetterPairs(str2.toUpperCase());
int intersection = 0;
int union = pairs1.size() + pairs2.size();
for (String s : pairs1) {
for (int j = 0; j < pairs2.size(); j++) {
if (s.equals(pairs2.get(j))) {
intersection++;
pairs2.remove(j);
break;
}
}
}
return (2.0 * intersection) / union;
}
private static List<String> wordLetterPairs(String str) {
List<String> AllPairs = new ArrayList<>();
String[] Words = str.split("\\s");
for (String word : Words) {
if (StringUtils.isNotBlank(word)) {
String[] PairsInWord = letterPairs(word);
Collections.addAll(AllPairs, PairsInWord);
}
}
return AllPairs;
}
private static String[] letterPairs(String str) {
int numPairs = str.length() - 1;
String[] pairs = new String[numPairs];
for (int i = 0; i < numPairs; i++) {
try {
pairs[i] = str.substring(i, i + 2);
} catch (Exception e) {
pairs[i] = str.substring(i, numPairs);
}
}
return pairs;
}
}

Answer 25

Here's another c++ implementation that follows the original article, that minimizes dynamic memory allocations.
It obtains the same matching values in the examples, but I think it's better to take into account also the single character words.
//---------------------------------------------------------------------------
// Similarity based on Sørensen–Dice index
double calc_similarity( const std::string_view s1, const std::string_view s2 )
{
// Check banal cases
if( s1.empty() || s2.empty() )
{// Empty string is never similar to another
return 0.0;
}
else if( s1==s2 )
{// Perfectly equal
return 1.0;
}
else if( s1.length()==1 || s2.length()==1 )
{// Single (not equal) characters have zero similarity
return 0.0;
}
/////////////////////////////////////////////////////////////////////////
// Represents a pair of adjacent characters
class charpair_t final
{
public:
charpair_t(const char a, const char b) noexcept : c1(a), c2(b) {}
[[nodiscard]] bool operator==(const charpair_t& other) const noexcept { return c1==other.c1 && c2==other.c2; }
private:
char c1, c2;
};
/////////////////////////////////////////////////////////////////////////
// Collects and access a sequence of adjacent characters (skipping spaces)
class charpairs_t final
{
public:
charpairs_t(const std::string_view s)
{
assert( !s.empty() );
const std::size_t i_last = s.size()-1;
std::size_t i = 0;
chpairs.reserve(i_last);
while( i<i_last )
{
// Accepting also single-character words (the second is a space)
//if( !std::isspace(s[i]) ) chpairs.emplace_back( std::tolower(s[i]), std::tolower(s[i+1]) );
// Skipping single-character words (as in the original article)
if( std::isspace(s[i]) ) ; // Skip
else if( std::isspace(s[i+1]) ) ++i; // Skip also next
else chpairs.emplace_back( std::tolower(s[i]), std::tolower(s[i+1]) );
++i;
}
}
[[nodiscard]] auto size() const noexcept { return chpairs.size(); }
[[nodiscard]] auto cbegin() const noexcept { return chpairs.cbegin(); }
[[nodiscard]] auto cend() const noexcept { return chpairs.cend(); }
auto erase(std::vector<charpair_t>::const_iterator i) { return chpairs.erase(i); }
private:
std::vector<charpair_t> chpairs;
};
charpairs_t chpairs1{s1},
chpairs2{s2};
const double orig_avg_bigrams_count = 0.5 * static_cast<double>(chpairs1.size() + chpairs2.size());
std::size_t matching_bigrams_count = 0;
for( auto ib1=chpairs1.cbegin(); ib1!=chpairs1.cend(); ++ib1 )
{
for( auto ib2=chpairs2.cbegin(); ib2!=chpairs2.cend(); )
{
if( *ib1==*ib2 )
{
++matching_bigrams_count;
ib2 = chpairs2.erase(ib2); // Avoid to match the same character pair multiple times
break;
}
else ++ib2;
}
}
return static_cast<double>(matching_bigrams_count) / orig_avg_bigrams_count;
}