我如何测量2个字符串之间的相似度?
给定两个字符串text1
和text2
public SOMEUSABLERETURNTYPE Compare(string text1, string text2)
{
// DO SOMETHING HERE TO COMPARE
}
例子:
第一个字符串:StackOverflow
第二个字符串:StaqOverflow
回报率:相似度为91%
回报可以在%或类似的东西。
第一个字符串:简单的文本测试
第二个字符串:复杂的文本测试
返回:这些值可以被认为是相等的
有任何想法吗? 做这个的最好方式是什么?
有多种不同的方式来做到这一点。 查看维基百科“字符串相似性度量”页面,了解链接到带算法的其他页面的链接。
然而,我认为这些算法都不考虑声音 - 所以“staq溢出”与“堆栈溢出”类似,尽管第一个在发音方面更为相似。
我刚刚发现了另一个页面,它提供了更多的选项......特别是,Soundex算法(维基百科)可能更接近你以后的内容。
Levenshtein距离可能是你正在寻找的。
以下是我为正在编写的项目编写的一些代码。 我需要知道字符串的相似比和基于字符串的字的相似比。 最后一个,我想知道最小字符串的词相似比(所以如果所有单词存在并匹配较大的字符串,结果将是100%)和较大字符串的字相似比(我称之为RealWordsRatio )。 我使用Levenshtein算法找到距离。 到目前为止,代码尚未优化,但它按预期工作。 希望对你有帮助。
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;
}
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