问题
Given two strings text1 and text2
public SOMEUSABLERETURNTYPE Compare(string text1, string text2)
{
// DO SOMETHING HERE TO COMPARE
}
Examples:
First String: StackOverflow
Second String: StaqOverflow
Return: Similarity is 91%
The return can be in % or something like that.
First String: The simple text test
Second String: The complex text test
Return: The values can be considered equal
Any ideas? What is the best way to do this?
回答1:
There are various different ways of doing this. Have a look at the Wikipedia "String similarity measures" page for links to other pages with algorithms.
I don't think any of those algorithms take sounds into consideration, however - so "staq overflow" would be as similar to "stack overflow" as "staw overflow" despite the first being more similar in terms of pronunciation.
I've just found another page which gives rather more options... in particular, the Soundex algorithm (Wikipedia) may be closer to what you're after.
回答2:
Levenshtein distance is probably what you're looking for.
回答3:
Here is some code I have written for a project I am working on. I need to know the Similarity Ratio of the strings and the Similarity Ratio based on words of the strings. This last one, I want to know both the Words Similarity Ratio of the smallest string(so if all words exist and match in the larger string the result will be 100%) and the Words Similarity Ratio of the larger string(which I call RealWordsRatio). I use the Levenshtein algorithm to find the distance. The code is unoptimised, so far, but it works as expected. I hope you find it useful.
public static int Compute(string s, string t)
{
int n = s.Length;
int m = t.Length;
int[,] d = new int[n + 1, m + 1];
// Step 1
if (n == 0)
{
return m;
}
if (m == 0)
{
return n;
}
// Step 2
for (int i = 0; i <= n; d[i, 0] = i++)
{
}
for (int j = 0; j <= m; d[0, j] = j++)
{
}
// Step 3
for (int i = 1; i <= n; i++)
{
//Step 4
for (int j = 1; j <= m; j++)
{
// Step 5
int cost = (t[j - 1] == s[i - 1]) ? 0 : 1;
// Step 6
d[i, j] = Math.Min(
Math.Min(d[i - 1, j] + 1, d[i, j - 1] + 1),
d[i - 1, j - 1] + cost);
}
}
// Step 7
return d[n, m];
}
double GetSimilarityRatio(String FullString1, String FullString2, out double WordsRatio, out double RealWordsRatio)
{
double theResult = 0;
String[] Splitted1 = FullString1.Split(new char[]{' '}, StringSplitOptions.RemoveEmptyEntries);
String[] Splitted2 = FullString2.Split(new char[]{' '}, StringSplitOptions.RemoveEmptyEntries);
if (Splitted1.Length < Splitted2.Length)
{
String[] Temp = Splitted2;
Splitted2 = Splitted1;
Splitted1 = Temp;
}
int[,] theScores = new int[Splitted1.Length, Splitted2.Length];//Keep the best scores for each word.0 is the best, 1000 is the starting.
int[] BestWord = new int[Splitted1.Length];//Index to the best word of Splitted2 for the Splitted1.
for (int loop = 0; loop < Splitted1.Length; loop++)
{
for (int loop1 = 0; loop1 < Splitted2.Length; loop1++) theScores[loop, loop1] = 1000;
BestWord[loop] = -1;
}
int WordsMatched = 0;
for (int loop = 0; loop < Splitted1.Length; loop++)
{
String String1 = Splitted1[loop];
for (int loop1 = 0; loop1 < Splitted2.Length; loop1++)
{
String String2 = Splitted2[loop1];
int LevenshteinDistance = Compute(String1, String2);
theScores[loop, loop1] = LevenshteinDistance;
if (BestWord[loop] == -1 || theScores[loop, BestWord[loop]] > LevenshteinDistance) BestWord[loop] = loop1;
}
}
for (int loop = 0; loop < Splitted1.Length; loop++)
{
if (theScores[loop, BestWord[loop]] == 1000) continue;
for (int loop1 = loop + 1; loop1 < Splitted1.Length; loop1++)
{
if (theScores[loop1, BestWord[loop1]] == 1000) continue;//the worst score available, so there are no more words left
if (BestWord[loop] == BestWord[loop1])//2 words have the same best word
{
//The first in order has the advantage of keeping the word in equality
if (theScores[loop, BestWord[loop]] <= theScores[loop1, BestWord[loop1]])
{
theScores[loop1, BestWord[loop1]] = 1000;
int CurrentBest = -1;
int CurrentScore = 1000;
for (int loop2 = 0; loop2 < Splitted2.Length; loop2++)
{
//Find next bestword
if (CurrentBest == -1 || CurrentScore > theScores[loop1, loop2])
{
CurrentBest = loop2;
CurrentScore = theScores[loop1, loop2];
}
}
BestWord[loop1] = CurrentBest;
}
else//the latter has a better score
{
theScores[loop, BestWord[loop]] = 1000;
int CurrentBest = -1;
int CurrentScore = 1000;
for (int loop2 = 0; loop2 < Splitted2.Length; loop2++)
{
//Find next bestword
if (CurrentBest == -1 || CurrentScore > theScores[loop, loop2])
{
CurrentBest = loop2;
CurrentScore = theScores[loop, loop2];
}
}
BestWord[loop] = CurrentBest;
}
loop = -1;
break;//recalculate all
}
}
}
for (int loop = 0; loop < Splitted1.Length; loop++)
{
if (theScores[loop, BestWord[loop]] == 1000) theResult += Splitted1[loop].Length;//All words without a score for best word are max failures
else
{
theResult += theScores[loop, BestWord[loop]];
if (theScores[loop, BestWord[loop]] == 0) WordsMatched++;
}
}
int theLength = (FullString1.Replace(" ", "").Length > FullString2.Replace(" ", "").Length) ? FullString1.Replace(" ", "").Length : FullString2.Replace(" ", "").Length;
if(theResult > theLength) theResult = theLength;
theResult = (1 - (theResult / theLength)) * 100;
WordsRatio = ((double)WordsMatched / (double)Splitted2.Length) * 100;
RealWordsRatio = ((double)WordsMatched / (double)Splitted1.Length) * 100;
return theResult;
}
回答4:
I wrote a Double Metaphone implementation in C# a while back. You'll find it vastly superior to Soundex and the like.
Levenshtein distance has also been suggested, and it's a great algorithm for a lot of uses, but phonetic matching is not really what it does; it only seems that way sometimes because phonetically similar words are also usually spelled similarly. I did an analysis of various fuzzy matching algorithms which you might also find useful.
回答5:
To deal with 'sound alikes' you may want to look into encoding using a phonetic algorithm like Double Metaphone or soundex. I don't know if computing Levenshtein distances on phonetic encoded strings would be beneficial or not, but might be a possibility. Alternately, you could use a heuristic like: convert each word in the string to its encoded form and remove any words that occur in both strings and replace them with a single representation before computing the Levenshtein distance.
回答6:
You might look for string "distances", for example the Levenshtein distance.
回答7:
Perl module Text::Phonetic has implementations of various algorithms.
回答8:
Jeff Atwood wrote about looking for a similar solution for determining the authorship of wiki posts which may help you narrow your search.
回答9:
If you're comparing values in a SQL database you can use the SOUNDEX function. If you query Google for SOUNDEX and C#, some people have written a similar function for that and VB.
回答10:
I have to recommend Soundex too, I have used it in the past to process misspelt city names. Here is a good link for usage: http://whitepapers.zdnet.com/abstract.aspx?docid=352953
回答11:
If you want to compare phonetically, check out the Soundex and Metaphone algorithms: http://www.blackbeltcoder.com/Articles/algorithms/phonetic-string-comparison-with-soundex
回答12:
Metaphone 3 is the third generation of the Metaphone algorithm. It increases the accuracy of phonetic encoding from the 89% of Double Metaphone to 98%, as tested against a database of the most common English words, and names and non-English words familiar in North America. This produces an extremely reliable phonetic encoding for American pronunciations.
Metaphone 3 was designed and developed by Lawrence Philips, who designed and developed the original Metaphone and Double Metaphone algorithms.
来源：https://stackoverflow.com/questions/1034622/how-can-i-measure-the-similarity-between-2-strings