[AccessD] Count the occurrences of each character in a string

[Arthur Fuller]

Thanks, Stuart,

Works like a charm.

On Mon, Sep 5, 2022 at 7:31 AM Stuart McLachlan <stuart at lexacorp.com.pg>
wrote:

> For a Unicode string with potentially thousands of code points, you don't
> want an array of all
> possible values.
>
> You can do it with an array the same size as the string length
> A bit more complicated than the previous code, but this one works well
> with either ANSI or
> Unicode strings.
>
>
> Function Lettercount(sInputString As String, CaseSensitive As Boolean) As
> Long
>     Dim lStr As Long, x As Long, y As Long, compare As Long
>     Dim sTxt As String
>     Dim freq() As Long
>
>     'Initialise
>     sTxt = sInputString 'we are going to modify the string!
>     If CaseSensitive = True Then
>         compare = vbBinaryCompare
>     Else
>         compare = vbTextCompare
>     End If
>     lStr = Len(sTxt)
>     ReDim freq(1 To lStr)
>
>     'Get the letter counts
>     For x = 1 To lStr
>        freq(x) = 1
>        For y = x + 1 To lStr
>            If StrComp(Mid$(sTxt, x, 1), Mid$(sTxt, y, 1), compare) = 0 Then
>                freq(x) = freq(x) + 1
>                Mid$(sTxt, y, 1) = Chr$(0)
>           End If
>        Next
>    Next
>
>   'Display the results
>    For x = 1 To lStr
>       If Mid$(sTxt, x, 1) <> Chr$(0) Then
>         Debug.Print Mid$(sTxt, x, 1) & "  -  " & str$(freq(x))
>       End If
>    Next
> End Function
>
>
>
>
> On 5 Sep 2022 at 19:29, Stuart McLachlan wrote:
>
> > Classic  "bucket" or "pigeon hole" application.
> >
> > For ANSI strings:
> >
> > DIm arr(32 to 255) 'assumes only printable values
> > for x = 1 to len(strTest)
> >    y = aSC(mid$(strTest,x,1)
> >   arr(y) = arr(y) + 1
> > next
> > for x = 32  to 255
> >    if arr(x) > 0 then  debug.print chr$(x) & " - " & str$(arr(x)) next
> >
> >
> > If it's a Wide String that can contain any Unicode point,  then you
> > really need to use a more complex solution such as a BTree or a
> > Collection where you check whether you already have an item with a key
> > matching the current character code and either add it or increment its
> > value.
> >
> >
> >
> > On 5 Sep 2022 at 4:43, Arthur Fuller wrote:
> >
> > > I'm trying to figure out the best way to count the occurrences of
> > > each character in a string. Sample string: "mississipi" I should get
> > > back an array looking like this:
> > >
> > > Character Count
> > > m 1
> > > i 4
> > > s 4
> > > p 1
> > >
> > > Any suggestions?
> > >
> > > --
> > > Arthur
> > > --
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> > > AccessD at databaseadvisors.com
> > > https://databaseadvisors.com/mailman/listinfo/accessd
> > > Website: http://www.databaseadvisors.com
> > >
> >
> >
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> >
>
>
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-- 
Arthur