Formula to fill down used range

[jazz2409]

Which calculates faster:

Range("G2").FormulaR1C1 = "=IFERROR(RC[-2]/RC[-1],0)"                
                With Sheets("comp")
                    .Range("G2").AutoFill .Range("G2:G" & .Cells(.Rows.Count, "A").End(xlUp).Row)
                End With

or

Lastrow = Range("A" & Rows.Count).End(xlUp).Row

Range("G2:G" & Lastrow).Formula ="=IFERROR(RC[-2]/RC[-1],0)"

and why?

Thank you!

[paulked]

You can put in a timer to test it (and any other routines) quite simply:

Sub Whatever()
    Dim tm as Double
    tm = Timer
'
' Your code
'
    Msgbox Timer - tm
End Sub

[jazz2409]

You can put in a timer to test it (and any other routines) quite simply:

Sub Whatever()
    Dim tm as Double
    tm = Timer
'
' Your code
'
    Msgbox Timer - tm
End Sub

Ooohh so this is what they are doing.. I've seen threads with people saying this ran for x seconds .. Thanks! I'll try now

[jazz2409]

You can put in a timer to test it (and any other routines) quite simply:

Sub Whatever()
    Dim tm as Double
    tm = Timer
'
' Your code
'
    Msgbox Timer - tm
End Sub

based on this, the second one works faster. Thank you!

[paulked]

You're welcome

It's a handy thing to use when testing

[jazz2409]

I actually included it in my project so I can check its performance anytime. Thank you.