hi

can you share a small sample data..

cheers!!


On Tue, May 6, 2014 at 10:59 PM, p178 <tdavi...@hotmail.com> wrote:

> Using Vbscript for an Excel 2010 worksheet, how can I find the start and
> end range of distinct values with a column ?
>
> I have a worksheet with some 200k rows, and 45 columns. 20 of such columns
> contain rows indicating a color value that I need to set. I have code that
> works but when working on 200k rows it is incredibly slow.
>
> Set objRange = objXLWs.UsedRange
> iRows = objRange.Rows.Count
> iColumns = objRange.Columns.Count
>
>                      For iR = 2 To iRows
>                  Select Case objRange.Item(iR, iC).Value
>                     Case "Y"
>                    objXLApp.Range(objRange.Item(
> iR, iC-1),objRange.Item(iR, iC-1)).Interior.ColorIndex = 6 'yellow
>                     Case "R"
>                    objXLApp.Range(objRange.Item(iR,
> iC-1),objRange.Item(iR, iC-1)).Interior.ColorIndex = 3 'red
>                     Case "O"
>                    objXLApp.Range(objRange.Item(iR,
> iC-1),objRange.Item(iR, iC-1)).Interior.ColorIndex = 45 'orange
>                     Case "T"
>                    objXLApp.Range(objRange.Item(iR,
> iC-1),objRange.Item(iR, iC-1)).Interior.ColorIndex = 42 'terquoise
>                     End Select
>                      Next
>
> What I am thinking is that sorting the column, bringing all the color
> values to the top, then finding the distinct ranges of each of the 4 x
> color values and trying to color by a range rather than individual row case
> statement. Sadly I have no idea how to do so. Obviously I would then resort
> by the original file format following all columns having been colored.
>
> Can anyone please give me some example code on how I would do this ? I
> have sort code that seems to work, I just cannot figure how to get start
> and end cell addresses for a block color statement.
>
> Set objRange2 = objXLWs.Cells(2, iC)
> objXLWs.UsedRange.Sort objRange2, xlAscending, , , , , , xlYes
>
> Many thanks for any help, Fin.
>
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