Forum Discussion
Index & Match Function
- Mar 31, 2025
Perhaps there are no data for 36 and 60 months ago?
Could you attach a small sample workbook demonstrating the problem (without sensitive data), or if that is not possible, make it available through OneDrive, Google Drive, Dropbox or similar? Alternatively, you can attach it to a private message to me. Thanks in advance.
Thank you, Hans. Problem solved.
However, I have another question. I am calculating standard deviation for a data set. I have tried the following two formulas on the same data set. However, I am getting different results. Any thoughts as to why? Thank you.
=STDEV.P(FILTER(D$48:D$437,$A$48:$A$437>EOMONTH($B7,-60)))*SQRT(12)
=STDEV.P(D146:D206)*SQRT(12)
Hi Richard,
That is impossible to answer without seeing a copy of the workbook.
- RichardR2060Apr 08, 2025Copper Contributor
Thank you very much for your help. See attached.
- HansVogelaarApr 08, 2025MVP
Thanks.
A12:A23 contains the dates in 2014.
$A$11:$A$400>EOMONTH($C3,-12) includes all dates in 2014 and later, so a much larger range.
If you want to filter for the year of the date in cell C3:
=STDEV.P(FILTER(C$11:C$400,YEAR($A$11:$A$400)=YEAR(C3)))*SQRT(12):
- RichardR2060Apr 08, 2025Copper Contributor
Thank you for your help. I am trying build a formula to calculate the standard deviation for the 10, 7, 5, 3 and 12 periods ended 03/31/2025. I came up with the below but don't think it is running the number correctly. Thank you again.
=STDEV.P(FILTER(C$12:C$178,$B$12:$B$178,EOMONTH($C1,-120)))*SQRT(12)