Question

我有一个for循环，它接受用户的输入和我的字典中的一个键并将它们传递给Damerau-Levenshtein函数并根据距离覆盖用户的输入字典键（for循环用于遍历每个字典键）。这对于大于三个字符的字符串足够好，但如果字符串是三个或更少字符，则算法返回错误的键。这是for循环：

    1950        For j = 0 To dict.Count - 1
    1960            distance = DamerauLevenshtein(SplitStr(i), dict.Keys(j))
    1970            'MsgBox dict.Keys(j) & vbCrLf & distance ' used for debugging
    1980            If distance < 4 Then
    1990                If distance < leastDist Then
    2000                    leastDist = distance
    2010                    SplitStr(i) = dict.Keys(j)
    2020                End If
    2030            End If
    2040        Next
    2050        MsgBox "The distance is: " & leastDist & vbCrLf & "The entered text was " & tempStr & vbCrLf & "The replaced word is " & SplitStr(i)

SplitStr（i）保存用户的输入，该输入来自分割功能。我任意选择了4个好距离

我从a bytes.com forum post偷走了算法。算法如下：

Function DamerauLevenshtein(str1, str2, Optional intSize = 256)
  Dim intTotalLen, arrDistance, intLen1, intLen2, i, j, arrStr1, arrStr2, arrDA, intMini
  Dim intDB, intI1, intJ1, intD

  str1 = UCase(str1)
  str2 = UCase(str2)
  intLen1 = Len(str1)
  intLen2 = Len(str2)
  intTotalLen = intLen1 + intLen2
  ReDim arrStr1(intLen1)
  ReDim arrStr2(intLen2)
  ReDim arrDA(intSize)
  ReDim arrDistance(intLen1 + 2, intLen2 + 2)
  arrDistance(0, 0) = intTotalLen

  For i = 0 To intSize - 1
      arrDA(i) = 0
  Next

  For i = 0 To intLen1
      arrDistance(i + 1, 1) = i
      arrDistance(i + 1, 0) = intTotalLen
  Next

  For i = 1 To intLen1
      arrStr1(i - 1) = Asc(Mid(str1, i, 1))
  Next

  For j = 0 To intLen2
      arrDistance(1, j + 1) = j
      arrDistance(0, j + 1) = intTotalLen
  Next

  For j = 1 To intLen2
      arrStr2(j - 1) = Asc(Mid(str2, j, 1))
  Next

  For i = 1 To intLen1
      intDB = 0

      For j = 1 To intLen2
          intI1 = arrDA(arrStr2(j - 1))
          intJ1 = intDB

          If arrStr1(i - 1) = arrStr2(j - 1) Then
              intD = 0
          Else
              intD = 1
          End If

          If intD = 0 Then intDB = j

          intMini = arrDistance(i, j) + intD
          If intMini > arrDistance(i + 1, j) + 1 Then intMini = arrDistance(i + 1, j) + 1
          If intMini > arrDistance(i, j + 1) + 1 Then intMini = arrDistance(i, j + 1) + 1
          If intMini > arrDistance(intI1, intJ1) + i - intI1 + j - intJ1 - 1 Then intMini = arrDistance(intI1, intJ1) + i - intI1 + j - intJ1 - 1

          arrDistance(i + 1, j + 1) = intMini
      Next

      arrDA(arrStr1(i - 1)) = i
  Next

  DamerauLevenshtein = arrDistance(intLen1 + 1, intLen2 + 1)
End Function

如果我输入＆＃34; Cire＆＃34;算法正确地返回＆＃34; CORE＆＃34;。

＆＃34;拉曼＆＃34;返回＆＃34; REMAN＆＃34; ＆＃34; Cosnigned＆＃34;返回＆＃34;委托

然而，＆＃34;现在＆＃34; 应该返回＆＃34;新＆＃34;但返回＆＃34; OCM＆＃34;。

＆＃34;新＆＃34;也回归＆＃34; OCM＆＃34; （所以距离应为0，但为2。）

＆＃34; FP＆＃34;应该是＆＃34; FP＆＃34;但返回＆＃34; OCM＆＃34;，距离是2

＆＃34; DPF＆＃34;应该是＆＃34; DPF＆＃34;但返回＆＃34; OCM＆＃34;，距离是2

我刚刚了解了这个算法，所以我确定我错过了一些重要的东西，但我无法看到它。想法？

Answer 1

我明白了。经过多次搜索，我发现一篇帖子说编辑距离通常为2.（他们没有说明为什么2是常见的任何优点）

我将if语句从4改为2，现在所有的问题条款都得到了纠正。

Damerau-Levenshtein算法不适用于短字符串

1 个答案: