VBA中的Levenshtein距离[闭合]

翻译自 Wikipedia :

Option Explicit
Public Function Levenshtein(s1 As String, s2 As String)

Dim i As Integer
Dim j As Integer
Dim l1 As Integer
Dim l2 As Integer
Dim d() As Integer
Dim min1 As Integer
Dim min2 As Integer

l1 = Len(s1)
l2 = Len(s2)
ReDim d(l1, l2)
For i = 0 To l1
    d(i, 0) = i
Next
For j = 0 To l2
    d(0, j) = j
Next
For i = 1 To l1
    For j = 1 To l2
        If Mid(s1, i, 1) = Mid(s2, j, 1) Then
            d(i, j) = d(i - 1, j - 1)
        Else
            min1 = d(i - 1, j) + 1
            min2 = d(i, j - 1) + 1
            If min2 < min1 Then
                min1 = min2
            End If
            min2 = d(i - 1, j - 1) + 1
            If min2 < min1 Then
                min1 = min2
            End If
            d(i, j) = min1
        End If
    Next
Next
Levenshtein = d(l1, l2)
End Function

?Levenshtein(“周六”、“周日”)

三

感谢smirkingman的代码贴。这是一个优化版本。

1)使用Asc(中间$(s1,i,1)代替。数值比较通常比文本快。

2)使用Mid$istead of Mid,因为后者是变体版本。加$就是字符串ver。

3)最少使用应用功能(仅限个人偏好)

4)使用Long而不是整数,因为这是excel本机使用的。

Function Levenshtein(ByVal string1 As String, ByVal string2 As String) As Long

Dim i As Long, j As Long
Dim string1_length As Long
Dim string2_length As Long
Dim distance() As Long

string1_length = Len(string1)
string2_length = Len(string2)
ReDim distance(string1_length, string2_length)

For i = 0 To string1_length
    distance(i, 0) = i
Next

For j = 0 To string2_length
    distance(0, j) = j
Next

For i = 1 To string1_length
    For j = 1 To string2_length
        If Asc(Mid$(string1, i, 1)) = Asc(Mid$(string2, j, 1)) Then
            distance(i, j) = distance(i - 1, j - 1)
        Else
            distance(i, j) = Application.WorksheetFunction.Min _
            (distance(i - 1, j) + 1, _
             distance(i, j - 1) + 1, _
             distance(i - 1, j - 1) + 1)
        End If
    Next
Next

Levenshtein = distance(string1_length, string2_length)

End Function

更新 :

对于那些想要的人:我认为可以肯定地说,大多数人使用Levenshtein距离来计算模糊匹配百分比。这是一种方法,我添加了一个优化,您可以指定要返回的最小匹配百分比(默认为70%+)。输入百分比,例如“50”或“80”,或“0”来运行公式。

速度的提高来自于这样一个事实:函数将通过检查两个字符串的长度来检查它是否可能在您给出的百分比内。请注意,有些地方可以优化此功能,但为了可读性,我将其保留在这里。为了证明功能性,我将距离串联起来,但您可以修改它:)

Function FuzzyMatch(ByVal string1 As String, _
                    ByVal string2 As String, _
                    Optional min_percentage As Long = 70) As String

Dim i As Long, j As Long
Dim string1_length As Long
Dim string2_length As Long
Dim distance() As Long, result As Long

string1_length = Len(string1)
string2_length = Len(string2)

' Check if not too long
If string1_length >= string2_length * (min_percentage / 100) Then
    ' Check if not too short
    If string1_length <= string2_length * ((200 - min_percentage) / 100) Then

        ReDim distance(string1_length, string2_length)
        For i = 0 To string1_length: distance(i, 0) = i: Next
        For j = 0 To string2_length: distance(0, j) = j: Next

        For i = 1 To string1_length
            For j = 1 To string2_length
                If Asc(Mid$(string1, i, 1)) = Asc(Mid$(string2, j, 1)) Then
                    distance(i, j) = distance(i - 1, j - 1)
                Else
                    distance(i, j) = Application.WorksheetFunction.Min _
                    (distance(i - 1, j) + 1, _
                     distance(i, j - 1) + 1, _
                     distance(i - 1, j - 1) + 1)
                End If
            Next
        Next
        result = distance(string1_length, string2_length) 'The distance
    End If
End If

If result <> 0 Then
    FuzzyMatch = (CLng((100 - ((result / string1_length) * 100)))) & _
                 "% (" & result & ")" 'Convert to percentage
Else
    FuzzyMatch = "Not a match"
End If

End Function

使用字节数组获得17倍的速度增益

  Option Explicit

  Public Declare Function GetTickCount Lib "kernel32" () As Long

  Sub test()
  Dim s1 As String, s2 As String, lTime As Long, i As Long
  s1 = Space(100)
  s2 = String(100, "a")
  lTime = GetTickCount
  For i = 1 To 100
     LevenshteinStrings s1, s2  ' the original fn from Wikibooks and Stackoverflow
  Next
  Debug.Print GetTickCount - lTime; " ms" '  3900  ms for all diff

  lTime = GetTickCount
  For i = 1 To 100
     Levenshtein s1, s2
  Next
  Debug.Print GetTickCount - lTime; " ms" ' 234  ms

  End Sub

  'Option Base 0 assumed

  'POB: fn with byte array is 17 times faster
  Function Levenshtein(ByVal string1 As String, ByVal string2 As String) As Long

  Dim i As Long, j As Long, bs1() As Byte, bs2() As Byte
  Dim string1_length As Long
  Dim string2_length As Long
  Dim distance() As Long
  Dim min1 As Long, min2 As Long, min3 As Long

  string1_length = Len(string1)
  string2_length = Len(string2)
  ReDim distance(string1_length, string2_length)
  bs1 = string1
  bs2 = string2

  For i = 0 To string1_length
      distance(i, 0) = i
  Next

  For j = 0 To string2_length
      distance(0, j) = j
  Next

  For i = 1 To string1_length
      For j = 1 To string2_length
          'slow way: If Mid$(string1, i, 1) = Mid$(string2, j, 1) Then
          If bs1((i - 1) * 2) = bs2((j - 1) * 2) Then   ' *2 because Unicode every 2nd byte is 0
              distance(i, j) = distance(i - 1, j - 1)
          Else
              'distance(i, j) = Application.WorksheetFunction.Min _
              (distance(i - 1, j) + 1, _
               distance(i, j - 1) + 1, _
               distance(i - 1, j - 1) + 1)
              ' spell it out, 50 times faster than worksheetfunction.min
              min1 = distance(i - 1, j) + 1
              min2 = distance(i, j - 1) + 1
              min3 = distance(i - 1, j - 1) + 1
              If min1 <= min2 And min1 <= min3 Then
                  distance(i, j) = min1
              ElseIf min2 <= min1 And min2 <= min3 Then
                  distance(i, j) = min2
              Else
                  distance(i, j) = min3
              End If

          End If
      Next
  Next

  Levenshtein = distance(string1_length, string2_length)

  End Function

我想它变得更快了。。。除了改进以前的代码速度和结果外,没有做太多其他的事情,比如%

' Levenshtein3 tweaked for UTLIMATE speed and CORRECT results
' Solution based on Longs
' Intermediate arrays holding Asc()make difference
' even Fixed length Arrays have impact on speed (small indeed)
' Levenshtein version 3 will return correct percentage
'
Function Levenshtein3(ByVal string1 As String, ByVal string2 As String) As Long

Dim i As Long, j As Long, string1_length As Long, string2_length As Long
Dim distance(0 To 60, 0 To 50) As Long, smStr1(1 To 60) As Long, smStr2(1 To 50) As Long
Dim min1 As Long, min2 As Long, min3 As Long, minmin As Long, MaxL As Long

string1_length = Len(string1):  string2_length = Len(string2)

distance(0, 0) = 0
For i = 1 To string1_length:    distance(i, 0) = i: smStr1(i) = Asc(LCase(Mid$(string1, i, 1))): Next
For j = 1 To string2_length:    distance(0, j) = j: smStr2(j) = Asc(LCase(Mid$(string2, j, 1))): Next
For i = 1 To string1_length
    For j = 1 To string2_length
        If smStr1(i) = smStr2(j) Then
            distance(i, j) = distance(i - 1, j - 1)
        Else
            min1 = distance(i - 1, j) + 1
            min2 = distance(i, j - 1) + 1
            min3 = distance(i - 1, j - 1) + 1
            If min2 < min1 Then
                If min2 < min3 Then minmin = min2 Else minmin = min3
            Else
                If min1 < min3 Then minmin = min1 Else minmin = min3
            End If
            distance(i, j) = minmin
        End If
    Next
Next

' Levenshtein3 will properly return a percent match (100%=exact) based on similarities and Lengths etc...
MaxL = string1_length: If string2_length > MaxL Then MaxL = string2_length
Levenshtein3 = 100 - CLng((distance(string1_length, string2_length) * 100) / MaxL)

End Function