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

[Arthur Fuller]

Hi Stuart,

I don't know what a PB function is. Also, I'm not sure where to place this
new code.

On Wed, Sep 7, 2022 at 3:13 AM Stuart McLachlan <stuart at lexacorp.com.pg>
wrote:

> Just converted this to a PB function and in doing so, I realised that it
> was slower than
> necessary. With a large string you will save a LOT of iterations with
>
> For x = 1 To lStr
>        If Asc(Mid$(sTxt, x, 1)) <> 0 Then
>             freq(x) = 1
>             For y = x + 1 To lStr
>                 If StrComp(Mid$(sTxt, x, 1), Mid$(sTxt, y, 1), compare) =
> 0 Then
>                     freq(x) = freq(x) + 1
>                     Mid$(sTxt, y, 1) = Chr$(0)
>                End If
>             Next
>        End If
> Next
>
>
> ...
>
>
> On 5 Sep 2022 at 21:31, Stuart McLachlan wrote:
>
> > For a Unicode string with potentially thousands of code points, you
> > don't want an array of all possible values.
> >
> > You can do it with an array the same size as the string length
> > A bit more complicated than the previous code, but this one works well
> >  with either ANSI or Unicode strings.
> >
> >
> > Function Lettercount(sInputString As String, CaseSensitive As Boolean)
> > As Long
> >     Dim lStr As Long, x As Long, y As Long, compare As Long
> >     Dim sTxt As String
> >     Dim freq() As Long
> >
> >     'Initialise
> >     sTxt = sInputString 'we are going to modify the string!
> >     If CaseSensitive = True Then
> >         compare = vbBinaryCompare
> >     Else
> >         compare = vbTextCompare
> >     End If
> >     lStr = Len(sTxt)
> >     ReDim freq(1 To lStr)
> >
> >     'Get the letter counts
> >     For x = 1 To lStr
> >        freq(x) = 1
> >        For y = x + 1 To lStr
> >            If StrComp(Mid$(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
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> > > > Website: http://www.databaseadvisors.com
> > > >
> > >
> > >
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> >
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-- 
Arthur