Question

我正在尝试使用Levenshtein Distance的帮助在OCR页面上找到模糊关键字（静态文本）。
为此，我想提供一定比例的允许错误（例如，15％）。

string Keyword = "past due electric service";

由于关键字长度为25个字符，我想允许4个错误（25 * .15向上舍入）
我需要能够将它与......进行比较。

string Entire_OCR_Page = "previous bill amount payment received on 12/26/13 thank 
                          you! current electric service total balances unpaid 7 
                          days after the total due date are subject to a late 
                          charge of 7.5% of the amount due or $2.00, whichever/5 
                          greater. "

这就是我现在正在做的事情......

int LevenshteinDistance = LevenshteinAlgorithm(Keyword, Entire_OCR_Page); // = 202   
int NumberOfErrorsAllowed = 4;   
int Allowance = (Entire_OCR_Page.Length() - Keyword.Length()) + NumberOfErrorsAllowed; // = 205

显然，Keyword中找不到OCR_Text（它不应该是{1}}。但是，使用Levenshtein的距离，误差的数量小于15％的余地（因此我的逻辑说它已被发现）。

有谁知道更好的方法吗？

Answer 1

使用子字符串回答了我的问题。发布以防其他人遇到相同类型的问题。有点不正统，但它对我很有用。

int TextLengthBuffer = (int)StaticTextLength - 1; //start looking for correct result with one less character than it should have.
int LowestLevenshteinNumber = 999999; //initialize insanely high maximum
decimal PossibleStringLength = (PossibleString.Length); //Length of string to search
decimal StaticTextLength = (StaticText.Length); //Length of text to search for
decimal NumberOfErrorsAllowed = Math.Round((StaticTextLength * (ErrorAllowance / 100)), MidpointRounding.AwayFromZero); //Find number of errors allowed with given ErrorAllowance percentage

    //Look for best match with 1 less character than it should have, then the correct amount of characters.
    //And last, with 1 more character. (This is because one letter can be recognized as 
    //two (W -> VV) and visa versa) 

for (int i = 0; i < 3; i++) 
{
    for (int e = TextLengthBuffer; e <= (int)PossibleStringLength; e++)
    {
        string possibleResult = (PossibleString.Substring((e - TextLengthBuffer), TextLengthBuffer));
        int lAllowance = (int)(Math.Round((possibleResult.Length - StaticTextLength) + (NumberOfErrorsAllowed), MidpointRounding.AwayFromZero));
        int lNumber = LevenshteinAlgorithm(StaticText, possibleResult);

        if (lNumber <= lAllowance && ((lNumber < LowestLevenshteinNumber) || (TextLengthBuffer == StaticText.Length && lNumber <= LowestLevenshteinNumber)))
        {
            PossibleResult = (new StaticTextResult { text = possibleResult, errors = lNumber });
            LowestLevenshteinNumber = lNumber;
        }
    }
    TextLengthBuffer++;
}




public static int LevenshteinAlgorithm(string s, string t) // Levenshtein Algorithm
{
    int n = s.Length;
    int m = t.Length;
    int[,] d = new int[n + 1, m + 1];

    if (n == 0)
    {
        return m;
    }

    if (m == 0)
    {
        return n;
    }

    for (int i = 0; i <= n; d[i, 0] = i++)
    {
    }

    for (int j = 0; 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;

            d[i, j] = Math.Min(
                Math.Min(d[i - 1, j] + 1, d[i, j - 1] + 1),
                d[i - 1, j - 1] + cost);
        }
    }
    return d[n, m];
}

Answer 2

我认为它不起作用，因为你的大量字符串是匹配的。所以我要做的就是尝试将你的关键字分成单独的单词。

然后找到OCR_TEXT中匹配这些单词的所有地方。

然后查看匹配的所有地方，看看这些地方中有4个是连续的并且与原始短语匹配。

我不确定我的解释是否清楚？

模糊匹配字符串中的多个单词

2 个答案: