'COUNTIFS w/ Multiple Criteria in Same Column
I'm trying to count instances of letters (like letters C through Z, excluding RR) within the same column if they are listed next to a name. Here's my sample Google Sheet that you can edit.
I'm trying to insert the formula in cell F3 that is highlighted yellow. So far I have...
=arrayformula(SUM(COUNTIFS(A2:A,{"C","D","E","F","G"},C2:C,E3:E)))
It seems like what I have should work, but it's giving me a #VALUE error, saying, "ARRAY arguments to COUNTIFS are of different size".
It seems like REGEXMATCH could be used inside the COUNTIF formula to make it easier to restrict the search to the range of letters I need, but not sure how to construct the formula.
Thanks for your help!
UPDATE This formula below works but only for column A. I actually need to specify a different range of letters to be counted in column B and totaled in column K.
=QUERY(A2:C,"select C,count(A) where A matches 'C|D|E|F|G|H|I|J|K' group by C label count(A)''", 0)
Seems like this post almost answers it.
Solution 1:[1]
A 2-formula solution was found to work as well. Answer is in the same spreadsheet linked above in Sheet2. It should prevent the first formula from not working should someone need a different one.
J2 - =QUERY(A2:C,"select C,count(A) where A matches 'C|D|E|F|G|H|I|J|K' group by C label count(A)''", 0)
L2 - =ArrayFormula(IFNA(vlookup(J2:J,QUERY(A2:C,"select C,count(B) where B matches 'F|G|H|I|J|K' group by C label count(B)''", 0),2,0)))
Solution 2:[2]
This formula should work for you if you paste it in cells F3 to F7. I currently have cells P1 to P26 (you'd probably want to hide this column) occupied with the letters of the alphabet in order, making it so that you can select P3:P26 (C-Z) to put into your first condition for your COUNTIFS.
=arrayformula(SUM(COUNTIFS(A2:A, $P$3:$P$26,C2:C,E3)))
You'll have to implement these formulas in the other places that you want them too, but it shouldn't be hard to change this formula to work in the other places as well.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Dharman |
| Solution 2 | Ihor Konovalenko |
