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 :thumb

It's a handy thing to use when testing :yes

[jazz2409]

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