Word Field needed to count number of words in a given bookmark - Wrong Answers

Posted on 2008-06-13
Last Modified: 2010-04-21
This is relative to

Working with GrahamSkan:

The routine is producing what appears to be the wrong answer for the following instance

In Word Edit>GoTo>Bookmark>Abstract in order to select the text being counted
In Immediate window ?ActiveDocument.Bookmarks("Abstract").Range gives [See the resulting text in snippet below]
In Immediate window ?selection.Words.Count gives  328
In Immediate window?ActiveDocument.Bookmarks("Abstract").Range.Words.Count gives 328
In Word the menu command Tools>Word Count gives 265
In Immediate window ?ubound(split(ActiveDocument.Bookmarks("Abstract").Range," ")) gives 270 (Origin is 0)
The correct answer is 270 manually counting and also counted in Excel to be 270.

What accounts for all the discrepancies?
328 from the routine;
265 from the builtin function;
270 by counting the words 3 other ways (manual, excel, VBA function Split)
I know - it's the ¿¿Twilight Zone?? Friday 13th, who knows.


Strategies for vascular control and limiting warm ischemia time vary between institutions for laparoscopic live donor nephrectomy (LLDN).  We refined our technique and retrospectively determined whether it safely provides an allograft of comparable quality to published series. Bla Bla Bla.

Patients and Methods:

Fifty consecutive LLDN between February 2003 and November 2006 were reviewed.  Key technical aspects include placing the perfused kidney and transected ureter entirely within an endocatch bag, with the string externalized through an extended lateral port site incision.  Vessels are then controlled with clips, or a Satinsky clamp for right sided veins.  The extraction incision is completed and the bag immediately withdrawn and placed on ice. Warm ischemia time (WIT) ends with perfusion with cold UW solution.


The series includes 42 left and 8 right kidneys.  13/50 (26%) demonstrated anatomical complexity (more than one artery, vein and/or ureter).  Average operative time was 178 minutes.  Average WIT was 129 seconds.  Conversion to open surgery occurred in 2 patients, one to define challenging anatomy, and another for hemorrhage from the renal artery stump.  Average blood loss was 120 ml.  Average length of stay was 3.6 days.

Average recipient creatinine was 1.26 mg/dl at discharge.  Delayed graft function occurred in three recipients.  ATN/slow normalization of creatinine occurred in four.  Ureteral injury occurred in one patient. Graft survival at 1 year was 96%.


The refined technique of LLDN mimics important principles of open donor nephrectomy.  Controllable variables which may impact graft function are optimized.  WIT is amongst the lowest reported for pure laparoscopy, without increasing complication rates, blood loss, or operative time.

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Question by:canesbr
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LVL 76

Expert Comment

ID: 21784621
As you can probably guess, the short answer is in the definition of what constitutes a Word.

The Words collection of ranges is created by splitting everything in its contained range into a word item. It seems that the splitting is done at a change from letters and numbers to something else. Trailing spaces are included with the item, but other characters are not, and thus become separate words.

This is obviously different from splitting at spaces.

I doubt that Microsoft publish the exact criteria for  the whole document word count in Tools or Properties, but the code in the snippet is gets closer to it.

I suspect that punctuation that could be embedded in a word such as a hyphenation characters and a dot when it is decimal point would account for some of the differences.
Function CountWords(rngText As Range) As Long

    Dim rngWord As Range

    Dim rngChar As Range

    For Each rngWord In rngText.words

        Set rngChar = rngWord.Characters.First

        Select Case rngChar.Text

            Case "0" To "9", "A" To "Z", "a" To "z"

                CountWords = CountWords + 1

        End Select

    Next rngWord

End Function

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LVL 76

Expert Comment

ID: 21784942
If you want to guarantee consistency, you could copy the relevant section to a temporary document and check the relevant document property.
Function CountWords2(rngText As Range) As Long

    Dim Doc As Document

    Set Doc = Documents.Add

    Doc.Range.Text = rngText.Text

    CountWords2 = Doc.BuiltInDocumentProperties(wdPropertyWords)

    Doc.Close wdDoNotSaveChanges

End Function

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Author Comment

ID: 21786884
Interesting (not so much)
"He/She was: 2.34"
I tried the same string a few different ways (see snippet)
3 words using function in Tools
Using the following method
Gives 9 if in one paragraph
Gives 20 if in a vertical stream
Gives 28 if each in a table

I guess that's what the documentation says, but  VBA nomenclature in MS-Word uses the word 'word' in a misleading way. Just fine to define a / as a word? I don't think so. An invisible paragraph mark as a word? Not so much. Call it a token.count  or something, but not a word.count.
Jeez Louize.
He/She was: 2.34								

Per Tools>Word Count&					

3 Words	 { He/She | was: | 2.34 }				

Per ActiveDocument.Bookmarks(MyBkmrk).Range.Words.Count 	

9 Words	 { He | / | She | was | : | 2 | . | 34 | ^p }

If in paragraphs					

20 Words							

	He	^p					

	/	^p					

	She	^p					


	Was	^p					

	:	^p					


	2	^p					

	.	^p					

	34	^p

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Author Comment

ID: 21786894
So it seems that there is no VB function that will replicate the Tools>Word Count..."
One has to use a kluge like creating a new doc?
LVL 76

Accepted Solution

GrahamSkan earned 500 total points
ID: 21788350
Yes as I said, it depends on what you mean by 'Word'.

This attempt is based on changing formatting characters to spaces and then splitting the text.

In my experiments it gives the same result as the document property, without the 'kludge' of creating a temporary document.
Function CountWords3(rngText As Range) As Long

    Dim strWords() As String

    Dim strText As String


    Set rngText = ActiveDocument.Range

    strText = Application.CleanString(rngText)

    strText = Replace(strText, vbCr, " ")

    strText = Replace(strText, vbVerticalTab, " ")

    strText = Replace(strText, vbTab, " ")

    Do While InStr(strText, "  ")

        strText = Replace(strText, "  ", " ")


    strText = Trim$(strText)

    If Len(strText) > 0 Then

        strWords = Split(strText, " ")

        CountWords3 = UBound(strWords) + 1

    End If

End Function

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Author Closing Comment

ID: 31467333
Also continue to feel that definition of word here is so divergent as to demand a different word, like token, as I said.
I find that a lot of solutions to my more tricky problems involve an iterative exchange (exactly). This one was particularly satisfying.

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