count unique with conditions

[\ditzman via OfficeKB.com\]

I am looking for a function or group of functions that can help me count the
unique entries that meet two other criteria. The table is similar to:

Column A: Column B: Column C: Column D:
12346 Level1 ACME 345334
12346 Level1 ACME 34523
12346 Level2 ACME 34622
12376 Level2 ACME 3453
13255 Level2 DODAD 345
13548 Level1 THINGABOB 322562
12356 Level1 WIDGET 34622
12376 Level2 WIDGET 3453
12378 Level3 WIDGET 546443

Column A is a list of unknown ID numbers. Column B&C values are known. The
table is sorted by Column C, then Column B, then Column A, all in acending
order. What I want is to be able to look for a value in Column C & Column B,
and then count the unique items in Column A that matches with Column B & C.

Example:
If I search for "ACME" & "LEVEL1" I should get a result of 1.
If I search for "ACME" & "LEVEL2" I should get a result of 2.

I do not need to know which ones are unique, I just need to know the total of
unique items as long as the line meets the other two criteria. I'm not sure
if this is even possible. I'm using Excel 2000.

Thanks for any help you can give me!!

--
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[Domenic]

Try...

=SUM(IF(FREQUENCY(IF((B1:B9="Level1")*(C1:C9="Acme"),A1:A9),IF((B1:B9="Le
vel1")*(C1:C9="Acme"),A1:A9))>0,1,0))

or

=COUNT(1/FREQUENCY(IF((B1:B9="Level1")*(C1:C9="Acme"),A1:A9),IF((B1:B9="L
evel1")*(C1:C9="Acme"),A1:A9)))

Both these formulas need to be confirmed with CONTROL+SHIFT+ENTER, not
just ENTER.

Hope this helps!

In article <[email protected]>,
\"ditzman via OfficeKB.com\" <[email protected]> wrote:

> I am looking for a function or group of functions that can help me count the
> unique entries that meet two other criteria. The table is similar to:
>
> Column A: Column B: Column C: Column D:
> 12346 Level1 ACME 345334
> 12346 Level1 ACME 34523
> 12346 Level2 ACME 34622
> 12376 Level2 ACME 3453
> 13255 Level2 DODAD 345
> 13548 Level1 THINGABOB 322562
> 12356 Level1 WIDGET 34622
> 12376 Level2 WIDGET 3453
> 12378 Level3 WIDGET 546443
>
> Column A is a list of unknown ID numbers. Column B&C values are known. The
> table is sorted by Column C, then Column B, then Column A, all in acending
> order. What I want is to be able to look for a value in Column C & Column B,
> and then count the unique items in Column A that matches with Column B & C.
>
> Example:
> If I search for "ACME" & "LEVEL1" I should get a result of 1.
> If I search for "ACME" & "LEVEL2" I should get a result of 2.
>
> I do not need to know which ones are unique, I just need to know the total of
> unique items as long as the line meets the other two criteria. I'm not sure
> if this is even possible. I'm using Excel 2000.
>
> Thanks for any help you can give me!!

[\ditzman via OfficeKB.com\]

Thanks! I believe it is working perfectly.

--
Message posted via http://www.officekb.com

[\ditzman via OfficeKB.com\]

I have one more question. If Column B was slightly modified:

Column A: Column B: Column C: Column D:
12346 Level1a ACME 345334
12346 Level1b ACME 34523
12346 Level2 ACME 34622
12376 Level2 ACME 3453
13255 Level2 DODAD 345
13548 Level1a THINGABOB 322562
12356 Level1b WIDGET 34622
12376 Level2 WIDGET 3453
12378 Level3 WIDGET 546443

Is there a way to do the same thing, but treat Level1a and Level1b as the
same? So I can search for Acme, and Level1a and Level1b and still come up
with a result of 1? I tried using "Level1*" in my function but it does not
like that.

Thanks again!

--
Message posted via http://www.officekb.com

[Domenic]

Try...

=SUM(IF(FREQUENCY(IF((LEFT(B1:B9,6)="Level1")*(C1:C9="Acme"),A1:A9),IF((L
EFT(B1:B9,6)="Level1")*(C1:C9="Acme"),A1:A9))>0,1,0))

or

=COUNT(1/FREQUENCY(IF((LEFT(B1:B9,6)="Level1")*(C1:C9="Acme"),A1:A9),IF((
LEFT(B1:B9,6)="Level1")*(C1:C9="Acme"),A1:A9)))

Both formulas need to be confirmed with CONTROL+SHIFT+ENTER.

Hope this helps!

In article <[email protected]>,
\"ditzman via OfficeKB.com\" <[email protected]> wrote:

> I have one more question. If Column B was slightly modified:
>
> Column A: Column B: Column C: Column D:
> 12346 Level1a ACME 345334
> 12346 Level1b ACME 34523
> 12346 Level2 ACME 34622
> 12376 Level2 ACME 3453
> 13255 Level2 DODAD 345
> 13548 Level1a THINGABOB 322562
> 12356 Level1b WIDGET 34622
> 12376 Level2 WIDGET 3453
> 12378 Level3 WIDGET 546443
>
> Is there a way to do the same thing, but treat Level1a and Level1b as the
> same? So I can search for Acme, and Level1a and Level1b and still come up
> with a result of 1? I tried using "Level1*" in my function but it does not
> like that.
>
> Thanks again!

[\ditzman via OfficeKB.com\]

That did it. Thanks again Domenic! It works great!

--
Message posted via http://www.officekb.com

[slc]

I have the similar situation except that Col A is not number but text. Tried both formulas (the ones without the LEFT function) but they returned 0 if Col A is text (not number). Can anyone help to advise how to amend the formula for this case. Thanks


Column A: Column B: Column C: Column D:
AP1 Level1a ACME 345334
AP1 Level1b ACME 34523
AP1 Level2 ACME 34622
AP2 Level2 ACME 3453
AP4 Level2 DODAD 345
AP3 Level1a THINGABOB 322562
AP5 Level1b WIDGET 34622
AP2 Level2 WIDGET 3453
AP6 Level3 WIDGET 546443

[Domenic]

Try...

=SUM(IF(FREQUENCY(IF((A1:A9<>"")*(LEFT(B1:B9,6)="Level1")*(C1:C9="Acme"),MATCH(A1:A9,A1:A9,0)),ROW(INDEX(A1:A9,0,0))-ROW(A1)+1)>0,1,0))

or

=COUNT(1/FREQUENCY(IF((A1:A9<>"")*(LEFT(B1:B9,6)="Level1")*(C1:C9="Acme"),MATCH(A1:A9,A1:A9,0)),ROW(INDEX(A1:A9,0,0))-ROW(A1)+1))

Both formulas need to be confirmed with CONTROL+SHIFT+ENTER, not just ENTER.

Hope this helps!

I have the similar situation except that Col A is not number but text. Tried both formulas (the ones without the LEFT function) but they returned 0 if Col A is text (not number). Can anyone help to advise how to amend the formula for this case. Thanks


Column A: Column B: Column C: Column D:
AP1 Level1a ACME 345334
AP1 Level1b ACME 34523
AP1 Level2 ACME 34622
AP2 Level2 ACME 3453
AP4 Level2 DODAD 345
AP3 Level1a THINGABOB 322562
AP5 Level1b WIDGET 34622
AP2 Level2 WIDGET 3453
AP6 Level3 WIDGET 546443

[slc]

Domenic,

Thanks.