Numbered column in vba

[PcMax]

Hi,

Seeking a simplified solution to achieve this result.
As follows:

From a list:

40850
40850
40850
40851
40851
40852
40852
40852
40852

what you need

3
2
1
2
1
4
3
2
1

Thank you in advance

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[Admin]

Hi

You could try this formula.

=COUNTIF($A$1:$A$9,A1)-COUNTIF($A$1:A1,A1)+1

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[PcMax]

Hi,

I messed around a bit 'to run in a loop vba.
I wonder if you can better manage this cycle with an array


Option Explicit
Public Const RigaUp = 2

Sub a()
Dim RigaDn As Long
Dim Giorni As Variant
Dim ciclo As Long
Dim ciclo1 As Long
Dim start As Integer
Dim Old As Integer
RigaDn = Cells(Rows.Count, 1).End(xlUp).Row
Giorni = Range(Cells(RigaUp, 4), Cells(RigaDn, 5)).Value
For ciclo = 1 To UBound(Giorni)
Giorni(ciclo, 1) = Int(Giorni(ciclo, 1))
Next
Range("AA2:AA21") = Giorni
For ciclo1 = 1 To UBound(Giorni)
Cells(ciclo1 + 1, 28) = Contig(Range("AA" & ciclo1 + 1 & ":AA22"), Int(Giorni(ciclo1, 1)))
Next
End Sub


Function Contig(ByRef Myaddr As Range, ByVal myVal As String) As Integer
For Each cell In Myaddr
If cell.Value = myVal Then
ccnt = ccnt + 1
Else
If ccnt > Contig Then Contig = ccnt
ccnt = 0
End If
Next cell
End Function

[Rick Rothstein]

Hi

You could try this formula.

=COUNTIF($A$1:$A$9,A1)-COUNTIF($A$1:A1,A1)+1
Another formula that can be used is this one...

=COUNTIF(A1:A$9,A1)

Change both A1 references to the cell reference for the cell that your data actually starts at.

[Rick Rothstein]

I messed around a bit 'to run in a loop vba.


If you really need to do this in VBA (instead of using one of the formulas Admin or I posted earlier), then you can use this macro...


Sub CountDupesBackwards()
Dim LastRow As Long
Const DataColumn As String = "A"
Const OutputColumn As String = "B"
Const StartRow As Long = 2
LastRow = Cells(Rows.Count, DataColumn).End(xlUp).Row
With Cells(StartRow, OutputColumn).Resize(LastRow - StartRow + 1)
.Formula = "=COUNTIF(" & DataColumn & StartRow & ":" & DataColumn & _
"$" & LastRow & "," & DataColumn & StartRow & ")"
.Value = .Value
End With
End Sub

[snb]

You could use:


Sub snb()
sn = Columns(1).SpecialCells(2)

For j = 1 To UBound(sn)
sn(j, 1) = UBound(Filter(Application.Transpose(sn), sn(j, 1))) + 1
Next

Columns(1).SpecialCells(2).Offset(, 3) = sn
End Sub

[PcMax]

Hi,

I have tried and tested various suggestions recommended.
Code Working successfully Thank Snb

[Rick Rothstein]

I have tried and tested various suggestions recommended.
Code Working successfully Thank Snb

One caution with snb's code... it will work fine as long as all the values in your list are the same length (as you showed in your posting). However, if the length of the list items can vary, and more importantly, if a smaller value can be found as part of a larger value (for example, if your list could have both 1850 and 41850 in it), then snb's code will return incorrect answers.

[snb]

Unless you adapt the code to that situation (which isn't hard to do)

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[PcMax]

Hi,

My research uses the data in a column that I find in columns and refer to a date for equity securities.
If all the data page are ordered and am in uniform, I should not have the problem reported.
If the dates were not in order, however, would get positioning errors of the formulas.
I could use an initial course to test the correctness of the data before running the code.