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Selection.Words.Count in Microsoft Word

I need a function that will give the count of words in a string that is equivalent to Word 2013's Selection.Words.Count property.  The behavior of the Word 2013 property appears to treat non-alphanumeric characters differently when more than 1 are in sequence.  The following shows selection in the left column, words.count in the second column:
Selection                   Words.Count      
Word                                      1      
.Word                                      2      
.Word.                                    3      Preceding and following periods count as words
Word/                                     2      Following / (and other non alphanumerics/non whitespace) count
Word/Second                        3      / counts as 1 word
Word//Second                       3      but 2 //s only count as 1 word
Word  Second                        2      2 spaces count as 1 delimiter (0 words)
Word<tab>Second               3      tab counts as a word
Word<tab><tab>Second     4      2 tabs count as 2 words
Word./Second                       3      period and slash count as 1 word
Word.Second                              3      1 period counts as 1 word
Word..                                    2      2 periods count as 1 word
Word…                                  2      3 periods count as 1 word
..Word                                    2      2 preceding periods count as 1 word
Word…Second                     3      3 periods between count as 1 word

Is there a simple way to duplicate this behavior with a string without having to use a selection in VBA or VB.Net?
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riverguy
Asked:
riverguy
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1 Solution
 
aikimarkCommented:
Since there is a Words collection on a range, why not use that instead of duplicating it in a user-defined function?  You could have a non-visible document used by your code to instantiate yous range object.  The Selection would not change.
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riverguyAuthor Commented:
Yes, I understand that I can do it that way.  But I need to call it from a procedure that uses a table look up on "tag" phases of up to 5 words each in fairly large documents, which already takes some time to run.  I was hoping to avoid the extra overhead of doing what you suggest.  I was thinking there is probably a function that Word uses itself and hoping to be able to call that function, or failing that reproduce it.  

Probably unnecessary but details of what I'm doing is locating each ":" in a document, then backing up 5 words, looking up in a table to determine presence in a table, and if found formatting the phrase (as a tag to highlight/categorize the following content), then if not found checking the previous 4 words, then 3, 2, 1.  Therefore I would need to call such a function many hundreds of time in a document.  Your suggestion is a fall back in case I can't identify or recreate a faster method.
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aikimarkCommented:
is this an approach to plagiarism detection?

I think you will get much better results if you state your actual problem, not just the obstacles you face with your chosen algorithm or approach.
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riverguyAuthor Commented:
No, not looking for plagiarism.  I'm categorizing content in medical records for identification and categorization of information such as reflexes, tenderness, medications, etc. in a way similar to what the Senate Watergate Committee staff used in analyzing voluminous records.  (I know, ancient history, but software was developed for that and I have used it.)  Tags, as defined in this app consist of 1 to 5 words, could be more, in a table.  I identify tags in the documents with a colon suffix, since it is frequently used for headings such as History,: Diagnosis:, etc, and many of the common tags used regularly in medical records overlap the set of tags that I am interested in for a particular case.  Rather than going through each of the several hundred Tags in the table and determining if the text preceding each colon contains that text, I have separated the tags using SQL into recordsets by word count, and starting with 5 words, I compare the preceding 5 words before each colon to lookup in the recordset for a match , then 4, 3, etc.  Since Words WordCount property uses word delimiters differently depending on if the delimiter is a space, hyphen, period, slash, etc. I wanted to store a word count that was identical to what Word does in the table.  Now I have already used a blank document for that whenever I add a new Tag to the table and that works.  And I know that I can compare the Tag with the rightmost portion of text preceding a colon, which I think would be less efficient, although frankly I haven't speed tested it.  But also partly from intellectual curiosity I wanted to find if anyone knew the algorithm that Word uses or if there is a an windows function that can call it.
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aikimarkCommented:
Is your text still in the database or have you extracted it to some flat files or Word documents?

It would be helpful to have some sample sentences/paragraphs/documents and the various phrases you are trying to detect.
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aikimarkCommented:
If you are going after the words, I would recommend using regular expressions.  Here are the 5-2 word parsing patterns before a colon.  These patterns are for the regexp object available in the VBA and VBScript environments.
(\w+)([^\w:]+)(\w+)([^\w:]+)(\w+)([^\w:]+)(\w+)([^\w:]+)(\w+) *:
(\w+)([^\w:]+)(\w+)([^\w:]+)(\w+)([^\w:]+)(\w+) *:
(\w+)([^\w:]+)(\w+)([^\w:]+)(\w+) *:
(\w+)([^\w:]+)(\w+) *:

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I tested the two word pattern against your question text example and got the following:
Match 0 Start(28) Length(14) 
SubMatch 0: Word
SubMatch 1: /
SubMatch 2: Second

Match 1 Start(44) Length(15) 
SubMatch 0: Word
SubMatch 1: //
SubMatch 2: Second

Match 2 Start(62) Length(13) 
SubMatch 0: Word
SubMatch 1: 
SubMatch 2: Second

Match 3 Start(77) Length(12) 
SubMatch 0: Word
SubMatch 1: 
SubMatch 2: Second

Match 4 Start(91) Length(13) 
SubMatch 0: Word
SubMatch 1: 
SubMatch 2: Second

Match 5 Start(106) Length(13) 
SubMatch 0: Word
SubMatch 1: ./
SubMatch 2: Second

Match 6 Start(121) Length(12) 
SubMatch 0: Word
SubMatch 1: .
SubMatch 2: Second

Match 7 Start(161) Length(12) 
SubMatch 0: Word
SubMatch 1: …
SubMatch 2: Second

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Note: The word separator does not have to be captured.  I included it so I could tell it was parsing correctly.

The Word patterns are a bit different.
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riverguyAuthor Commented:
Although the problem is solved I would still like to know if there is a way to call Microsoft's function used in returning the Words.Count property.
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aikimarkCommented:
...if there is a way to call Microsoft's function used in returning the Words.Count property
I'm not sure I understand.

The title of this question implies that you already know how to get to the Selection.Words.Count property.

Is the question one of run-time environment VBA inside Word vs Word automation (VBScript, Excel, Access, Powershell, etc) or something else?
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riverguyAuthor Commented:
Something else.  I want to call it independently of Microsoft Word, as from vb.net but without having to open a word document, insert the text into a new selection and then retrieve the word count.  I haven't yet tested the regex you supplied, but I need it to exactly replicate the Words.count property in Word.  I suspect that somewhere there is available an API call that can be used.  I guess It's a matter of curiosity, since I can get the same results with a word document, but if anyone is familiar with the internal function that Microsoft Uses and it is accessible, I'm interested in that.  But thanks for all  your help.
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aikimarkCommented:
If you have a reliable source of word parsing, relative to a trailing colon character, why do you need to exactly replicate what Word does?

Have you looked at the OpenXML SDK?
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riverguyAuthor Commented:
When there are periods, hyphens and other punctuation characters, Word handles them differently as I showed in the sample.  If the word count I apply to a phrase (that I have previously determined with my own function) does not match exactly the way Word does in say .movestart wdWords,-5 and my Non word function treats the word count differently a phrase that should match doesn't.  Now my question is solved by inserting the phrase in a selection in a blank document in the background and getting Words.WordCount.  So the answer to your question of why I need to exactly replicate what Word does is to that all wordphrases I have assigned a 5 count to will match backing up 5 words in the target document selection.  
Beyond that I am simply curious to know if I can access Word's function directly.  Bottom line is curiosity!!  Yes, I know that's dumb!
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aikimarkCommented:
I think this is the Word VBA sequence that selects five "words", including the trailing colon character:
selection.Find.Execute ":"
selection.Moveright wdcharacter, 1, false
selection.MoveLeft wdword, 5, true

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To test this, I typed the following into the Word document:
Now is the time for all good men: to come to the aid of their country.
After executing the above statements, I checked the results in the immediate window
?"""" & selection.Text & """"
"for all good men:"
?selection.Words.Count
 5 
for w=1 to selection.Words.Count: ?w,"""" & selection.Words(w) & """":next
 1            "for "
 2            "all "
 3            "good "
 4            "men"
 5            ": "

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Note: Even though the selected text ended with a colon, the words collection in the selection included the trailing space.

As far as I can tell, you can't directly execute the Word functions without instantiating Word and opening the document.

While I appreciate your curiosity, I think the best solution would be to look for the word sequences ignoring the punctuation.  Forget what Word does.  Parse the text.
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