What is the best way to compare 2 strings to see how similar they are?
Examples:
My String
My String With Extra Words
Or
static class LevenshteinDistance
{
public static int Compute(string s, string t)
{
if (string.IsNullOrEmpty(s))
{
if (string.IsNullOrEmpty(t))
return 0;
return t.Length;
}
if (string.IsNullOrEmpty(t))
{
return s.Length;
}
int n = s.Length;
int m = t.Length;
int[,] d = new int[n + 1, m + 1];
// initialize the top and right of the table to 0, 1, 2, ...
for (int i = 0; i <= n; d[i, 0] = i++);
for (int j = 1; j <= m; d[0, j] = j++);
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
int cost = (t[j - 1] == s[i - 1]) ? 0 : 1;
int min1 = d[i - 1, j] + 1;
int min2 = d[i, j - 1] + 1;
int min3 = d[i - 1, j - 1] + cost;
d[i, j] = Math.Min(Math.Min(min1, min2), min3);
}
}
return d[n, m];
}
}
If anyone was wondering what the C# equivalent of what @FrankSchwieterman posted is:
public static int GetDamerauLevenshteinDistance(string s, string t)
{
if (string.IsNullOrEmpty(s))
{
throw new ArgumentNullException(s, "String Cannot Be Null Or Empty");
}
if (string.IsNullOrEmpty(t))
{
throw new ArgumentNullException(t, "String Cannot Be Null Or Empty");
}
int n = s.Length; // length of s
int m = t.Length; // length of t
if (n == 0)
{
return m;
}
if (m == 0)
{
return n;
}
int[] p = new int[n + 1]; //'previous' cost array, horizontally
int[] d = new int[n + 1]; // cost array, horizontally
// indexes into strings s and t
int i; // iterates through s
int j; // iterates through t
for (i = 0; i <= n; i++)
{
p[i] = i;
}
for (j = 1; j <= m; j++)
{
char tJ = t[j - 1]; // jth character of t
d[0] = j;
for (i = 1; i <= n; i++)
{
int cost = s[i - 1] == tJ ? 0 : 1; // cost
// minimum of cell to the left+1, to the top+1, diagonally left and up +cost
d[i] = Math.Min(Math.Min(d[i - 1] + 1, p[i] + 1), p[i - 1] + cost);
}
// copy current distance counts to 'previous row' distance counts
int[] dPlaceholder = p; //placeholder to assist in swapping p and d
p = d;
d = dPlaceholder;
}
// our last action in the above loop was to switch d and p, so p now
// actually has the most recent cost counts
return p[n];
}