Use AVERAGEIF with "OR" expression?

Thanks for replying so quickly, SG. Unfortunately that doesn’t solve my issue but I may not have been totally clear explaining it.

In the solution you offer, if a single “Friday”, “Saturday” or “Sunday” is found in row 29, the AVERAGE function would average all cells in row 26, even those without corresponding values in row 29 of “Friday”, ”Saturday” or “Sunday” (e.g., even those that have “Monday” or “Tuesday” or any other value) and that’s unfortunately not what I’m trying to do.

My table is a “life log”, so each column is a day of the year and each row is something I’m tracking (e.g., hours spent exercising, hours worked, alcoholic beverages consumed, etc.). I have an autopopulated cell in each column for which day of the week it is, and in a separate graph, I am trying to average certain values for certain types of days of the week (hence “Friday”/“Saturday”/“Sunday”). I need an AVERAGEIF(...) or AVERAGEIFS(...) that can help me achieve that and thereby give me an accurate average for, say, number of alcoholic beverages on those days, or hours worked on those days, etc.

I don’t think using OR(...) within the AVERAGEIF function will work due to limitations with booleans and I’m trying to avoid overly complex solutions such as the ones mentioned in the first post, or even another option like averaging Fridays, Saturdays and Sunday separately from each other in additional placeholder fields and then averaging those three results together. So I’m hoping I can achieve the result I’m looking for with one function in a cell, without adding more rows/cells/tables. Hopefully that’s not impossible in Numbers.

Thanks again for any help with this issue!

Thanks Wayne, I could do that and get a sum of the averages, but even if I averaged those averages by dividing that sum by three, that wouldn’t be an exactly accurate average — e.g., imagine that my table has many fewer Sundays than Fridays or Saturdays - say there’s only one ”Sunday” and the value in the row I’m concerned with for that Sunday column is “300” but all the Friday and Saturday columns, of which there are thousands in this extreme example, have “0” in every cell in that row, in this example, the proposed solution would create an average of “100” [300 divided by 3] even though the actual average is near zero [300 divided by, say, 3,000]).

That’s an extreme example and in my case, your solution would be closer to the right answer but would still be wrong and I’m looking for exact numbers. Is there any way to pull this off?

Thanks again!

Thank you so much! I was unaware of the COUNTIF(...) function and your solution absolutely solves my problem. Thanks for the very fast response and great answer and sorry that I wasn't as fast thanking for you as I should've been.