我正在使用此代码进行句子类似,代码在java上可用我想在c#中使用它。
public static int getWordChanges(String s1, String s2) {
int similarityThreshold = 50;
int wordChanges = 0;
s1 = s1.toLowerCase().replace(".", "").replace(",", "").replace(";", "");
s2 = s2.toLowerCase().replace(".", "").replace(",", "").replace(";", "");
//Loop through each word in s1
for (int i = 0; i < s1.split(" ").length; i++) {
boolean exists = false;
//Search for i'th word in s1 in s2
for (int j = 0; j < s2.split(" ").length; j++) {
//Is the word misspelled?
if ((getLevenshteinDistance(s1.split(" ")[i], s2.split(" ")[j]) * 100 / s1.split(" ")[i].length()) < similarityThreshold) {
exists = true;
break;
}
}
//If the word does not exist, increment wordChanges
if (!exists) {
wordChanges++;
}
}
return wordChanges;
}
这是我希望在c#中执行此代码的Java代码 转换c#
中的代码后 public int getWordChanges(String s1, String s2)
{
int similarityThreshold = 50;
int wordChanges = 0;
s1 = s1.ToLower().Replace(".", "").Replace(",", "").Replace(";", "");
s2 = s2.ToLower().Replace(".", "").Replace(",", "").Replace(";", "");
//Loop through each word in s1
for (int i = 0; i < s1.Split(' ').Length; i++)
{
bool exists = false;
//Search for i'th word in s1 in s2
for (int j = 0; j < s2.Split(' ').Length; j++)
{
//Is the word misspelled?
if ((getLevenshteinDistance(s1.Split(' ')[i], s2.Split(' ')[j]) * 100 / s1.Split(' ')[i].Length()) < similarityThreshold)
{
exists = true;
break;
}
}
//If the word does not exist, increment wordChanges
if (!exists)
{
wordChanges++;
}
}
return wordChanges;
}
}
}
此行有错误
if ((getLevenshteinDistance(s1.Split(' ')[i], s2.Split(' ')[j]) * 100 / s1.Split(' ')[i].Length()) < similarityThreshold)
长度错误将显示我如何解决这个问题
答案 0 :(得分:1)
将此功能添加到您的项目
test-test-test@2016-10-04.txt#
并更改public static int getLevenshteinDistance(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];
}
因为String.Length是属性而不是方法