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I found this bit of code that computes Levenshtein's distance between an answer and a guess:
int CheckErrors(string Answer, string Guess)
{
int[,] d = new int[Answer.Length + 1, Guess.Length + 1];
for (int i = 0; i <= Answer.Length; i++)
d[i, 0] = i;
for (int j = 0; j <= Guess.Length; j++)
d[0, j] = j;
for (int j = 1; j <= Guess.Length; j++)
for (int i = 1; i <= Answer.Length; i++)
if (Answer[i - 1] == Guess[j - 1])
d[i, j] = d[i - 1, j - 1]; //no operation
else
d[i, j] = Math.Min(Math.Min(
d[i - 1, j] + 1, //a deletion
d[i, j - 1] + 1), //an insertion
d[i - 1, j - 1] + 1 //a substitution
);
return d[Answer.Length, Guess.Length];
}
But I need a way to do a count for the amount of times each error occurs. Is there an easy way to implement that?
798
The Levenshtein distance is usually calculated by preparing a matrix of size (M+1)x(N+1) —where M and N are the lengths of the 2 words—and looping through said matrix using 2 for loops, performing some calculations within each iteration.
The Levenshtein distance used as a metric provides a boost to accuracy of an NLP model by verifying each named entity in the entry. The vector search solution does a good job, and finds the most similar entry as defined by the vectorization.
Levenshtein distance is a lexical similarity measure which identifies the distance between one pair of strings. It does so by counting the number of times you would have to insert, delete or substitute a character from string 1 to make it like string 2.
The Levenshtein distance (a.k.a edit distance) is a measure of similarity between two strings. It is defined as the minimum number of changes required to convert string a into string b (this is done by inserting, deleting or replacing a character in string a ).
Seems like you could add counters for each of the operations:
if (Answer[i - 1] == Guess[j - 1])
d[i, j] = d[i - 1, j - 1]; //no operation
else
{
int del = d[i-1, j] + 1;
int ins = d[i, j-1] + 1;
int sub = d[i-1, j-1] + 1;
int op = Math.Min(Math.Min(del, ins), sub);
d[i, j] = op;
if (i == j)
{
if (op == del)
++deletions;
else if (op == ins)
++insertions;
else
++substitutions;
}
}
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