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Scott is looking for a way to get a "frequency count" of all the characters in a document. He would like to know how many times each character, ASCII codes 9 through 255, occur. It is possible to use Find and Replace to determine the count of individual characters (simply search for a character in question and then replace it with itself), but such an approach would be tedious, at best, if you needed to do it for 247 different character codes to get the desired information.
Such a task must be done with a macro, but there are several ways to approach it. One way is to write a quick macro that will step through each member of the Character collection, examining each, and assigning that character to one of a number of counters.
Sub CountChars1() Dim iCount(0 To 255) As Integer Dim i As Integer Dim vCharacter As Variant Dim sTemp As String ' Initialize the array For i = 0 To 255 iCount(i) = 0 Next i ' Fill the array For Each oCharacter In ActiveDocument.Characters i = Asc(oCharacter) iCount(i) = iCount(i) + 1 Next ' Add document for results Documents.Add Selection.TypeText Text:="ASCII Character Count" & vbCrLf ' Only output codes 9 through 255 For i = 9 To 255 sTemp = Chr(i) If i < 32 Then sTemp = Trim(Str(i)) sTemp = sTemp & Chr(9) & Trim(Str(iCount(i))) sTemp = sTemp & vbCrLf Selection.TypeText Text:=sTemp Next i End Sub
The macro uses the iCount array to accumulate the counts of each character code, and then a new document is created to output the results. (The results document can be formatted in any way desired.)
This approach can work well for relatively short documents, up to a few pages. When the document gets longer, the macro gets slower. Why? Because it takes a great deal of time to use the Characters collection for some reason. If the macro runs too slow for your documents, then you will want to change it a bit so that it works solely with strings.
Sub CountChars2() Dim iCount(0 To 255) As Long Dim i As Long Dim j as integer Dim lCharCount As Long Dim sDoc As String Dim sTemp As String ' Initialize the array For i = 0 To 255 iCount(i) = 0 Next i ' Assign document to a huge string lCharCount = ActiveDocument.Characters.Count sDoc = ActiveDocument.Range(0, lCharCount) ' Fill the array For i = 1 to Len(sDoc) j = Asc(Mid(sDoc, i, 1)) iCount(j) = iCount(j) + 1 Next ' Add document for results Documents.Add Selection.TypeText Text:="ASCII Character Count" & vbCrLf ' Only output codes 9 through 255 For i = 9 To 255 sTemp = Chr(i) If i < 32 Then sTemp = Trim(Str(i)) sTemp = sTemp & Chr(9) & Trim(Str(iCount(i))) sTemp = sTemp & vbCrLf Selection.TypeText Text:=sTemp Next i End Sub
Notice that this version of the macro stuffs the entire document into a single string, sDoc. This string can then be processed very, very quickly by the macro. (A 635-page document only took about 30 seconds to process on my system.) Because this version is made to work with longer documents, note as well that some of the variable types have been changed to reflect the likelihood of larger counts.
WordTips is your source for cost-effective Microsoft Word training. (Microsoft Word is the most popular word processing software in the world.) This tip (112) applies to MS Word versions: 2007 | 2010
You can find a version of this tip for the older menu interface of Word here: Character Frequency Count.
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