excel - Distancia de Levenshtein en VBA

excel-vba levenshtein-distance

Question

Tengo una hoja de Excel con datos que quiero obtener Distancia Levenshtein entre ellos. Ya traté de exportar como texto, leer desde script (php), ejecutar Levenshtein (calcular Distancia de Levenshtein), guardarlo para sobresalir de nuevo.

Pero estoy buscando una manera de calcular programáticamente una distancia de Levenshtein en VBA. ¿Cómo voy a hacerlo?

Answer 1

Creo que fue aún más rápido ... No hice mucho más que mejorar el código anterior para la velocidad y los resultados como%

'' 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

Answer 2

Gracias a Smirkingman por la buena publicación de código. Aquí hay una versión optimizada.

1) Use Asc (Mid $ (s1, i, 1) en su lugar. La comparación numérica es generalmente más rápida que el texto.

2) Use Mid $ istead of Mid ya que la última es la variante ver. y agregar $ es string ver.

3) Use la función de aplicación para mín. (preferencia personal solamente)

4) Usa Long en vez de enteros, ya que es lo que Excel usa de manera nativa.

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

ACTUALIZAR :

Para aquellos que lo quieren: creo que es seguro decir que la mayoría de la gente usa la distancia de Levenshtein para calcular porcentajes de coincidencia difusa. Aquí hay una manera de hacerlo, y he agregado una optimización para que pueda especificar el mínimo. match% to return (valor predeterminado es 70% +. Ingrese porcentajes como "50" u "80", o "0" para ejecutar la fórmula independientemente).

El aumento de velocidad proviene del hecho de que la función verificará si es posible incluso que esté dentro del porcentaje que le das verificando la longitud de las 2 cuerdas. Tenga en cuenta que hay algunas áreas donde se puede optimizar esta función, pero la he mantenido en este punto por razones de legibilidad. Concatené la distancia en el resultado como prueba de funcionalidad, pero puedes jugar con ella :)

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

Answer 3

Traducido de 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 ("sábado", "domingo")

3

Answer 4

Usa una matriz de bytes para obtener un aumento de velocidad de 17x

 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
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