¿Alguna vez has necesitado contar el número de palabras en una cadena?
Exploremos algunas formas en las que podemos lograrlo.
Uso de VBA simple (más o menos, pero no del todo)
A continuación se muestran 2 funciones que se pueden emplear para obtener el recuento de palabras de una cadena. Sin embargo, esto solo funciona correctamente si se sigue la estructura de oración adecuada. Además, ambos enfoques se basan en el uso de mi función TrimMultipleSpaces() que, a su vez, se basa en RegEx de todos modos. Por eso, le recomendaría que simplemente use uno de los 2 enfoques RegEx presentados en la siguiente sección, ya que son más precisos/confiables y solo realizan una llamada RegEx singular.
'---------------------------------------------------------------------------------------
' Procedure : CountWords
' Author : Daniel Pineault, CARDA Consultants Inc.
' Website :
' Purpose : Counts the number of word in a string.
' Good for basic usage, but RegEx/MSHTML approaches are more reliable
' Copyright : The following is release as Attribution-ShareAlike 4.0 International
' (CC BY-SA 4.0) -
' Req'd Refs: None required
' Dependencies: TrimMultipleSpaces()
'
'
' Input Variables:
' ~~~~~~~~~~~~~~~~
' sInput : String to count the words of
'
' Usage:
' ~~~~~~
' ? CountWords(" Just a quick test !?")
' Returns -> 4
'
' ? CountWords(" Just a quick test !?Another sentence would go here.")
' Returns -> 9
'
' ? CountWords(" Just a quick test !?Another sentence would go here .")
' Returns -> 10 '********incorrect
'
' Revision History:
' Rev Date(yyyy-mm-dd) Description
' **************************************************************************************
' 1 2024-03-23
'---------------------------------------------------------------------------------------
Public Function CountWords(ByVal vInput As Variant) As Long
On Error GoTo Error_Handler
If Len(Trim(vInput & vbNullString)) = 0 Then GoTo Error_Handler_Exit
vInput = TrimMultipleSpaces(vInput) 'Correct for multiple spaces
vInput = Trim(vInput)
CountWords = UBound(Split(vInput, " ")) + 1
Error_Handler_Exit:
On Error Resume Next
Exit Function
Error_Handler:
MsgBox "The following error has occurred" & vbCrLf & vbCrLf & _
"Error Source: CountWords" & vbCrLf & _
"Error Number: " & Err.Number & vbCrLf & _
"Error Description: " & Err.Description & _
Switch(Erl = 0, "", Erl 0, vbCrLf & "Line No: " & Erl) _
, vbOKOnly + vbCritical, "An Error has Occurred!"
Resume Error_Handler_Exit
End Function'---------------------------------------------------------------------------------------
' Procedure : CountWords
' Author : Daniel Pineault, CARDA Consultants Inc.
' Website :
' Purpose : Word count that ignores punctuation
' Good for basic usage, but RegEx/MSHTML approaches are more reliable
' Copyright : The following is release as Attribution-ShareAlike 4.0 International
' (CC BY-SA 4.0) -
' Req'd Refs: None required
' Dependencies: TrimMultipleSpaces()
'
'
' Input Variables:
' ~~~~~~~~~~~~~~~~
' sInput : String to count the words of
'
' Usage:
' ~~~~~~
' ? CountWords(" Just a quick test !?")
' Returns -> 4
'
' ? CountWords(" Just a quick test !?Another sentence would go here.")
' Returns -> 9
'
' ? CountWords(" Just a quick test !?Another sentence would go here .")
' Returns -> 10 '********incorrect
'
' Revision History:
' Rev Date(yyyy-mm-dd) Description
' **************************************************************************************
' 1 2024-03-23
'---------------------------------------------------------------------------------------
Public Function CountWords(ByVal vInput As Variant) As Long
On Error GoTo Error_Handler
Dim lCharsLeft As Long
If Len(Trim(vInput & vbNullString)) = 0 Then GoTo Error_Handler_Exit
vInput = TrimMultipleSpaces(vInput) 'Correct for multiple spaces
vInput = Trim(vInput)
lCharsLeft = Len(Replace(vInput, " ", ""))
If lCharsLeft = 0 Then
CountWords = 0
Else
CountWords = Len(vInput) - Len(Replace(vInput, " ", "")) + 1
End If
Error_Handler_Exit:
On Error Resume Next
Exit Function
Error_Handler:
MsgBox "The following error has occurred" & vbCrLf & vbCrLf & _
"Error Source: CountWords" & vbCrLf & _
"Error Number: " & Err.Number & vbCrLf & _
"Error Description: " & Err.Description & _
Switch(Erl = 0, "", Erl 0, vbCrLf & "Line No: " & Erl) _
, vbOKOnly + vbCritical, "An Error has Occurred!"
Resume Error_Handler_Exit
End Function¡Usando el poder de las expresiones regulares!
Dado que las dos funciones anteriores requerían expresiones regulares y no eran precisas en todos los casos, exploré la posibilidad de usar expresiones regulares puras para lograrlo. ¡En expresiones regulares, es increíblemente fácil obtener un recuento de palabras!
Solo necesitamos usar el patrón simple ‘\w+(-\w+)*’ y luego contar la cantidad de coincidencias encontradas. Es muy sencillo. RegEx hace todo el trabajo pesado y se ocupa de todo lo demás por nosotros.
Uso de la biblioteca RegEx de VBScript
'---------------------------------------------------------------------------------------
' Procedure : RegEx_CountWords
' Author : Daniel Pineault, CARDA Consultants Inc.
' Website :
' Purpose : Count the number of word in a string. (ignores punctuation)
' Copyright : The following is release as Attribution-ShareAlike 4.0 International
' (CC BY-SA 4.0) -
' Req'd Refs: Late Binding -> None required
' Early Binding -> Microsoft VBScript Regular Expressions 5.5
'
' Input Variables:
' ~~~~~~~~~~~~~~~~
' sInput : String to count the words of
'
' Usage:
' ~~~~~~
' ? RegEx_CountWords(" Just a quick test !?")
' Returns -> 4
'
' ? RegEx_CountWords(" Just a quick test !?Another sentence would go here.")
' Returns -> 9
'
' Revision History:
' Rev Date(yyyy-mm-dd) Description
' **************************************************************************************
' 1 2024-03-23
'---------------------------------------------------------------------------------------
Public Function RegEx_CountWords(ByVal vInput As Variant) As Long
On Error GoTo Error_Handler
#Const RegEx_EarlyBind = False 'True => Early Binding / False => Late Binding
#If RegEx_EarlyBind = True Then
Dim oRegEx As VBScript_RegExp_55.RegExp
Dim oMatches As VBScript_RegExp_55.MatchCollection
Set oRegEx = New VBScript_RegExp_55.RegExp
#Else
Dim oRegEx As Object
Dim oMatches As Object
Set oRegEx = CreateObject("VBScript.RegExp")
#End If
If Not IsNull(vInput) Then
With oRegEx
.Pattern = "\w+(-\w+)*" 'Remove multi-space and leading/trailing spaces
.Global = True
.IgnoreCase = True
.MultiLine = True
Set oMatches = .Execute(vInput)
End With
RegEx_CountWords = oMatches.Count
Else
RegEx_CountWords = 0
End If
Error_Handler_Exit:
On Error Resume Next
Set oMatches = Nothing
Set oRegEx = Nothing
Exit Function
Error_Handler:
MsgBox "The following error has occured" & vbCrLf & vbCrLf & _
"Error Number: " & Err.Number & vbCrLf & _
"Error Source: RegEx_CountWords" & vbCrLf & _
"Error Description: " & Err.Description & _
Switch(Erl = 0, "", Erl 0, vbCrLf & "Line No: " & Erl) _
, vbOKOnly + vbCritical, "An Error has Occured!"
Resume Error_Handler_Exit
End FunctionUso de la biblioteca MSHTML RegEx
Planificando con un poco de anticipación cuándo se retirarían las expresiones regulares de VBScript, también desarrollé una versión alternativa utilizando la biblioteca de objetos HTML de Microsoft.
'---------------------------------------------------------------------------------------
' Procedure : MSHTML_CountWords
' Author : Daniel Pineault, CARDA Consultants Inc.
' Website :
' Purpose : Count the number of word in a string. (ignores punctuation)
' Version of RegEx using MSHTML to get around VBScript deprecation
' Copyright : The following is release as Attribution-ShareAlike 4.0 International
' (CC BY-SA 4.0) -
' Req'd Refs: Late Binding -> None required
' Early Binding -> Microsoft HTML Object Library
'
' Input Variables:
' ~~~~~~~~~~~~~~~~
' sInput : String to count the words of
'
' Usage:
' ~~~~~~
' ? MSHTML_CountWords(" Just a quick test !?")
' Returns -> 4
'
' ? MSHTML_CountWords(" Just a quick test !?Another sentence would go here.")
' Returns -> 9
'
' Revision History:
' Rev Date(yyyy-mm-dd) Description
' **************************************************************************************
' 1 2024-03-23
'---------------------------------------------------------------------------------------
Public Function MSHTML_CountWords(ByVal vInput As Variant) As Long
On Error GoTo Error_Handler
#Const HTMLFile_EarlyBind = False 'True => Early Binding / False => Late Binding
#If HTMLFile_EarlyBind = True Then
Dim oHTMLFile As MSHTML.HTMLDocument
Dim oElem As MSHTML.HTMLGenericElement
Set oHTMLFile = New MSHTML.HTMLDocument
#Else
Dim oHTMLFile As Object
Dim oElem As Object
Set oHTMLFile = CreateObject("HTMLFile")
#End If
If Len(Trim(vInput & vbNullString)) = 0 Then GoTo Error_Handler_Exit
oHTMLFile.body.innerHTML = "" 'Clear the default line
Set oElem = oHTMLFile.createElement("p")
Call oElem.setAttribute("id", "result")
Call oHTMLFile.body.appendChild(oElem)
Set oElem = oHTMLFile.createElement("script")
Call oElem.setAttribute("type", "text/javascript") 'Not strictly necessary as this is the default value anyways
oElem.innerText = "function regEx_WordCount() {" & _
" var regEx_Pattern = /\w+(-\w+)*/igm;" & _
" var myInput="" & vInput & "";" & _
" var regEx_Matches = myInput.match(regEx_Pattern);" & _
" document.getElementById('result').innerText = regEx_Matches.length;" & _
" return regEx_Matches.length;" & _
"}"
Call oHTMLFile.body.appendChild(oElem)
Call oHTMLFile.parentWindow.execScript("regEx_WordCount();")
MSHTML_CountWords = oHTMLFile.getElementById("result").innerText
Error_Handler_Exit:
On Error Resume Next
Set oElem = Nothing
Set oHTMLFile = Nothing
Exit Function
Error_Handler:
MsgBox "The following error has occurred" & vbCrLf & vbCrLf & _
"Error Source: MSHTML_CountWords" & vbCrLf & _
"Error Number: " & Err.Number & vbCrLf & _
"Error Description: " & Err.Description & _
Switch(Erl = 0, "", Erl 0, vbCrLf & "Line No: " & Erl) _
, vbOKOnly + vbCritical, "An Error has Occurred!"
Resume Error_Handler_Exit
End FunctionUso de la automatización de palabras
'---------------------------------------------------------------------------------------
' Procedure : Word_CountWords
' Author : Daniel Pineault, CARDA Consultants Inc.
' Website :
' Purpose : Count the number of word in a string via Word automation
' Copyright : The following is release as Attribution-ShareAlike 4.0 International
' (CC BY-SA 4.0) -
' Req'd Refs: Late Binding -> None required
' Early Binding -> Microsoft Word XX.X Object Library
'
' Input Variables:
' ~~~~~~~~~~~~~~~~
' sInput : String to count the words of
'
' Usage:
' ~~~~~~
' ? Word_CountWords(" Just a quick test !?")
' Returns -> 5 !!!WRONG!!! Come one Microsoft.
'
' ? Word_CountWords(" Just a quick test !?Another sentence would go here.")
' Returns -> 9
'
' Revision History:
' Rev Date(yyyy-mm-dd) Description
' **************************************************************************************
' 1 2024-03-23
'---------------------------------------------------------------------------------------
Public Function Word_CountWords(ByVal sInput As String) As Long
On Error GoTo Error_Handler
#Const Word_EarlyBind = False 'True => Early Binding / False => Late Binding
'Microsoft Word XX.X Object Library
#If Word_EarlyBind = True Then
Dim oWord As Word.Application
Dim oDoc As Word.Document
Set oWord = New Word.Application
#Else
Dim oWord As Object
Dim oDoc As Object
Const wdStatisticWords = 0
Set oWord = CreateObject("Word.Application")
#End If
oWord.Visible = False
Set oDoc = oWord.Documents.Add
oDoc.Activate
oWord.Selection.TypeText sInput
'Word_CountWords = oDoc.Words.Count - 1 'Incorrect value that take into account irrelevant stuff! KB 212705
Word_CountWords = oDoc.ComputeStatistics(wdStatisticWords)
Error_Handler_Exit:
On Error Resume Next
oWord.Quit False
Set oWord = Nothing
Exit Function
Error_Handler:
oWord.Visible = True ' Don't leave a hidden instance!
MsgBox "The following error has occurred" & vbCrLf & vbCrLf & _
"Error Source: Word_CountWords" & vbCrLf & _
"Error Number: " & Err.Number & vbCrLf & _
"Error Description: " & Err.Description & _
Switch(Erl = 0, "", Erl 0, vbCrLf & "Line No: " & Erl) _
, vbOKOnly + vbCritical, "An Error has Occurred!"
Resume Error_Handler_Exit
End FunctionLamentablemente, el recuento que devuelve Word puede ser incorrecto. Consulte el siguiente artículo para obtener más detalles:
Error en Microsoft Word: el recuento de palabras es incorrecto
Mientras trabajaba en mi último artículo, también desarrollé una función de automatización de Word para utilizar el contador de palabras integrado de Microsoft Word. Pensé que podría haber sido la solución más sencilla para obtener un recuento de palabras rápido y confiable de cualquier cadena. Estaba muy equivocado.
Así que ahí lo tienes. Al final del día, tienes dos funciones RegEx confiables que puedes usar para obtener el recuento de palabras en cualquier cadena. Más allá de eso, te recomendaría no usar VBA simple o automatización de Word.
