我正在创建一个程序来检查单词是否是简化单词（txt，msg等），如果是简化的，它会找到正确的拼写，如txt = text，msg = message。我在c＃中使用了NHunspell建议方法，它提出了所有可能的结果。
问题是如果我输入“txt”，结果是text，tat，tot等。我不知道如何选择正确的单词。我使用了Levenshtein距离（C# - Compare String Similarity）但结果仍然是1。
输入：txt 结果：text = 1，ext = 1 tit = 1
你能帮我解释如何获得简化单词的含义或拼写正确吗？ 示例：msg
答案 0 :(得分：6)
我使用您的示例数据测试了您的输入，只有text的距离为25而另一个的距离为33.这是我的代码：
string input = "TXT";
string[] words = new[]{"text","tat","tot"};
var levenshtein = new Levenshtein();
const int maxDistance = 30;
var distanceGroups = words
.Select(w => new
{
Word = w,
Distance = levenshtein.iLD(w.ToUpperInvariant(), input)
})
.Where(x => x.Distance <= maxDistance)
.GroupBy(x => x.Distance)
.OrderBy(g => g.Key)
.ToList();
foreach (var topCandidate in distanceGroups.First())
Console.WriteLine("Word:{0} Distance:{1}", topCandidate.Word, topCandidate.Distance);
这是levenshtein类：
public class Levenshtein
{
///*****************************
/// Compute Levenshtein distance
/// Memory efficient version
///*****************************
public int iLD(String sRow, String sCol)
{
int RowLen = sRow.Length; // length of sRow
int ColLen = sCol.Length; // length of sCol
int RowIdx; // iterates through sRow
int ColIdx; // iterates through sCol
char Row_i; // ith character of sRow
char Col_j; // jth character of sCol
int cost; // cost
/// Test string length
if (Math.Max(sRow.Length, sCol.Length) > Math.Pow(2, 31))
throw (new Exception("\nMaximum string length in Levenshtein.iLD is " + Math.Pow(2, 31) + ".\nYours is " + Math.Max(sRow.Length, sCol.Length) + "."));
// Step 1
if (RowLen == 0)
{
return ColLen;
}
if (ColLen == 0)
{
return RowLen;
}
/// Create the two vectors
int[] v0 = new int[RowLen + 1];
int[] v1 = new int[RowLen + 1];
int[] vTmp;
/// Step 2
/// Initialize the first vector
for (RowIdx = 1; RowIdx <= RowLen; RowIdx++)
{
v0[RowIdx] = RowIdx;
}
// Step 3
/// Fore each column
for (ColIdx = 1; ColIdx <= ColLen; ColIdx++)
{
/// Set the 0'th element to the column number
v1[0] = ColIdx;
Col_j = sCol[ColIdx - 1];
// Step 4
/// Fore each row
for (RowIdx = 1; RowIdx <= RowLen; RowIdx++)
{
Row_i = sRow[RowIdx - 1];
// Step 5
if (Row_i == Col_j)
{
cost = 0;
}
else
{
cost = 1;
}
// Step 6
/// Find minimum
int m_min = v0[RowIdx] + 1;
int b = v1[RowIdx - 1] + 1;
int c = v0[RowIdx - 1] + cost;
if (b < m_min)
{
m_min = b;
}
if (c < m_min)
{
m_min = c;
}
v1[RowIdx] = m_min;
}
/// Swap the vectors
vTmp = v0;
v0 = v1;
v1 = vTmp;
}
// Step 7
/// Value between 0 - 100
/// 0==perfect match 100==totaly different
///
/// The vectors where swaped one last time at the end of the last loop,
/// that is why the result is now in v0 rather than in v1
//System.Console.WriteLine("iDist=" + v0[RowLen]);
int max = System.Math.Max(RowLen, ColLen);
return ((100 * v0[RowLen]) / max);
}
///*****************************
/// Compute the min
///*****************************
private int Minimum(int a, int b, int c)
{
int mi = a;
if (b < mi)
{
mi = b;
}
if (c < mi)
{
mi = c;
}
return mi;
}
///*****************************
/// Compute Levenshtein distance
///*****************************
public int LD(String sNew, String sOld)
{
int[,] matrix; // matrix
int sNewLen = sNew.Length; // length of sNew
int sOldLen = sOld.Length; // length of sOld
int sNewIdx; // iterates through sNew
int sOldIdx; // iterates through sOld
char sNew_i; // ith character of sNew
char sOld_j; // jth character of sOld
int cost; // cost
/// Test string length
if (Math.Max(sNew.Length, sOld.Length) > Math.Pow(2, 31))
throw (new Exception("\nMaximum string length in Levenshtein.LD is " + Math.Pow(2, 31) + ".\nYours is " + Math.Max(sNew.Length, sOld.Length) + "."));
// Step 1
if (sNewLen == 0)
{
return sOldLen;
}
if (sOldLen == 0)
{
return sNewLen;
}
matrix = new int[sNewLen + 1, sOldLen + 1];
// Step 2
for (sNewIdx = 0; sNewIdx <= sNewLen; sNewIdx++)
{
matrix[sNewIdx, 0] = sNewIdx;
}
for (sOldIdx = 0; sOldIdx <= sOldLen; sOldIdx++)
{
matrix[0, sOldIdx] = sOldIdx;
}
// Step 3
for (sNewIdx = 1; sNewIdx <= sNewLen; sNewIdx++)
{
sNew_i = sNew[sNewIdx - 1];
// Step 4
for (sOldIdx = 1; sOldIdx <= sOldLen; sOldIdx++)
{
sOld_j = sOld[sOldIdx - 1];
// Step 5
if (sNew_i == sOld_j)
{
cost = 0;
}
else
{
cost = 1;
}
// Step 6
matrix[sNewIdx, sOldIdx] = Minimum(matrix[sNewIdx - 1, sOldIdx] + 1, matrix[sNewIdx, sOldIdx - 1] + 1, matrix[sNewIdx - 1, sOldIdx - 1] + cost);
}
}
// Step 7
/// Value between 0 - 100
/// 0==perfect match 100==totaly different
//System.Console.WriteLine("Dist=" + matrix[sNewLen, sOldLen]);
int max = System.Math.Max(sNewLen, sOldLen);
return (100 * matrix[sNewLen, sOldLen]) / max;
}
}
答案 1 :(得分：0)
不是一个完整的解决方案，只是一个有希望的有用的建议...
在我看来，人们不太可能使用与正确单词一样长的简化，因此您至少可以过滤掉长度<=输入长度的所有结果。
答案 2 :(得分：0)
扩展我的评论，您可以使用正则表达式搜索输入的“扩展”结果。像这样：
private int stringSimilarity(string input, string result)
{
string regexPattern = ""
foreach (char c in input)
regexPattern += input + ".*"
Match match = Regex.Match(result, regexPattern,
RegexOptions.IgnoreCase);
if (match.Success)
return 1;
else
return 0;
}
忽略1和0 - 我不知道相似性的估值是如何起作用的。
答案 3 :(得分：0)
您确实需要实现SQL中存在的SOUNDEX例程。我已经在以下代码中完成了这个：
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace Soundex
{
class Program
{
static char[] ignoreChars = new char[] { 'a', 'e', 'h', 'i', 'o', 'u', 'w', 'y' };
static Dictionary<char, int> charVals = new Dictionary<char, int>()
{
{'b',1},
{'f',1},
{'p',1},
{'v',1},
{'c',2},
{'g',2},
{'j',2},
{'k',2},
{'q',2},
{'s',2},
{'x',2},
{'z',2},
{'d',3},
{'t',3},
{'l',4},
{'m',5},
{'n',5},
{'r',6}
};
static void Main(string[] args)
{
Console.WriteLine(Soundex("txt"));
Console.WriteLine(Soundex("text"));
Console.WriteLine(Soundex("ext"));
Console.WriteLine(Soundex("tit"));
Console.WriteLine(Soundex("Cammmppppbbbeeelll"));
}
static string Soundex(string s)
{
s = s.ToLower();
StringBuilder sb = new StringBuilder();
sb.Append(s.First());
foreach (var c in s.Substring(1))
{
if (ignoreChars.Contains(c)) { continue; }
// if the previous character yields the same integer then skip it
if ((int)char.GetNumericValue(sb[sb.Length - 1]) == charVals[c]) { continue; }
sb.Append(charVals[c]);
}
return string.Join("", sb.ToString().Take(4)).PadRight(4, '0');
}
}
}
请参阅此代码，您提供的示例中唯一匹配的内容是text。运行控制台应用程序，您将看到输出（即txt将匹配text）。
答案 4 :(得分：0)
我认为像单词用来纠正拼写的程序的一种方法是使用NLP（自然语言处理）技术来获得在拼写错误的上下文中使用的名词/形容词的顺序。然后将其与已知的句子结构进行比较他们可以估计70％的拼写错误是名词的可能性，并使用该信息来过滤纠正的拼写。
SharpNLP看起来像一个很好的图书馆，但我还没有机会摆弄它。为了构建一个已知句子结构库BTW，在uni中我们将算法应用于公共领域书籍。
结帐sams simMetrics library我在SO（download here，docs here）上找到了加载Levenshtein距离以外算法的更多选项。