Excel/Vba How to conditionally change the number format of ALL workbook cells with numbers in them, based on their number values?

I think this is a unique question which I need help with.

This code does NOT work but I hope there is a syntax that would work and I would also need to know where to put it.

In which event and at what level of code (Module or ThisWorkbook or Sheet)?

Or can this be done in Excel with some type of Conditional Formatting?

If Cells.Value = 0 Then
    Cells.NumberFormat = "0"
ElseIf Abs(Cells.Value) > 0 And Abs(Cells.Value) < 0.1 Then
    Cells.NumberFormat = "0.000"
ElseIf Abs(Cells.Value) >= 0.1 And Abs(Cells.Value) < 1 Then
    Cells.NumberFormat = "0.00"
ElseIf Abs(Cells.Value) >= 1 And Abs(Cells.Value) < 10 Then
    Cells.NumberFormat = "0.0"
ElseIf Abs(Cells.Value) >= 10 Then
    Cells.NumberFormat = "0"
End If

Thanks, Dale

EDIT: My current issue with trying to use Conditional Formatting:

Solution

Format Cells with Numbers Conditionally in the Whole Workbook

This is a tool that will format the cells with numbers as required.
Before using it, create a copy of your file to test it.
Copy the complete code into a standard module e.g. Module1 of the workbook.
Run the code and determine if that is what you need.
You will have to save the file using either the .xlsm or .xlsb extension to keep the code in the workbook.
The code took about 25 seconds for 100k cells with numbers on a decade old machine. Share how long it took on your side.

Option Explicit

Sub FormatCellsWithNumbers()

    Const SECONDS_PER_100k As Long = 25 ' depends on the computer
    Const WARNING_CELLS_COUNT As Long = 100000
    
    Dim wb As Workbook: Set wb = ThisWorkbook ' workbook containing this code
    ' If it isn't, reference it by its name or use 'ActiveWorkbook'.

    ' Count the number of cells with numbers.
    Dim CellsCount As Double: CellsCount = CellsWithNumbersCount(wb)
    If CellsCount > WARNING_CELLS_COUNT Then
        Dim msg As Long: msg = MsgBox("Found " & Format(CellsCount, "#,##0") _
            & " cells with numbers." & vbLf _
            & "Formatting that many cells will take about " _
            & Format(CellsCount / 100000 * SECONDS_PER_100k, "#,##0") _
            & " seconds." & vbLf & vbLf & "Do you want to continue?", _
            vbQuestion + vbYesNo + vbDefaultButton2, "Long Operation!")
        If msg = vbNo Then Exit Sub
    End If
    
    ' Start measuring the time passed.
    Dim t As Double: t = Timer
    
    Application.ScreenUpdating = False ' increase efficiency
    
    ' Declare additional variables.
    Dim ws As Worksheet, cell As Range, FormattedCellsCount As Long
    
    ' Loop through all worksheets and format cells containing numbers.
    For Each ws In wb.Worksheets
        For Each cell In ws.UsedRange.Cells
            FormatCellWithNumber cell, FormattedCellsCount
        Next cell
    Next ws

    Application.ScreenUpdating = True

    ' Inform.
    MsgBox Format(FormattedCellsCount, "#,##0") _
        & " cells with numbers formatted in " _
        & Format(Timer - t, "0") & " seconds.", vbInformation

End Sub

Function CellsWithNumbersCount(ByVal wb As Workbook) As Double
    
    Dim ws As Worksheet, cell As Range, CellsCount As Long
    
    For Each ws In wb.Worksheets
        For Each cell In ws.UsedRange.Cells
            If VarType(cell.Value) = vbDouble Then CellsCount = CellsCount + 1
        Next cell
    Next ws
    
    CellsWithNumbersCount = CellsCount
    
End Function

Sub FormatCellWithNumber(ByVal cell As Range, ByRef FormattedCellsCount As Long)
    
    Dim Value As Variant: Value = cell.Value
    
    Dim NumFormat As String
    
    If VarType(Value) = vbDouble Then ' is a number
        FormattedCellsCount = FormattedCellsCount + 1
        Select Case Abs(Value)
            Case 0, Is >= 10: NumFormat = "0"
            Case Is < 0.1: NumFormat = "0.000"
            Case Is < 1: NumFormat = "0.00"
            Case Is < 10: NumFormat = "0.0"
        End Select
        cell.NumberFormat = NumFormat
    'Else ' not a number
    End If

End Sub