Counting unique combinations in two columns
 

Hi,

I am interested in finding a function that will count the number of
unique combinations a in two columns (columns A& B). For example:

Column A Column B Column C Column D
ABC 03/2007 =A1&B1 =SUMPRODUCT((1/
COUNTIF(C1:C5,C1:C5&"")))
BCA 03/2007 =A2&B2
CAB 07/2007 =A3&B3
ACB 07/2007 =A4&B4
ABC 03/2007 =A5&B5

In the case above, there are 4 unique combinations in columns A and B
(ABC 03/2007, BCA 03/2007, CAB 07/2007, ACB 07/2007). I don't count
the second instance of ABC 03/2007 becuase it is already in the first
row.

Is there a cleaner way of doing this? I've tried using sumproduct, but
I can't seem to get it to work the way I want it to. Does anyone know
how this can be done?

There are only four possible strings in column A (ABC, BCA, CAB, ACB)

Thanks!
Jason

Counting unique combinations in two columns
 

Providing you have no blanks in your concatenated range try:-

=SUM(1/COUNTIF(C1:C5,C1:C5))

If you have blanks (which will cause your formula below to fail) then try:-

=SUM(IF(FREQUENCY(IF(LEN(C1:C5)0,MATCH(C1:C5,C1:C 5,0),""),IF(LEN(C1:C5)0,MATCH(C1:C5,C1:C5,0),"")) 0,1))

Both are array formula ctrl+shift+enter

Mike


" wrote:

Hi,

I am interested in finding a function that will count the number of
unique combinations a in two columns (columns A& B). For example:

Column A Column B Column C Column D
ABC 03/2007 =A1&B1 =SUMPRODUCT((1/
COUNTIF(C1:C5,C1:C5&"")))
BCA 03/2007 =A2&B2
CAB 07/2007 =A3&B3
ACB 07/2007 =A4&B4
ABC 03/2007 =A5&B5

In the case above, there are 4 unique combinations in columns A and B
(ABC 03/2007, BCA 03/2007, CAB 07/2007, ACB 07/2007). I don't count
the second instance of ABC 03/2007 becuase it is already in the first
row.

Is there a cleaner way of doing this? I've tried using sumproduct, but
I can't seem to get it to work the way I want it to. Does anyone know
how this can be done?

There are only four possible strings in column A (ABC, BCA, CAB, ACB)

Thanks!
Jason

Counting unique combinations in two columns
 

You could also try
=SUMPRODUCT(--(MATCH(A1:A5&B1:B5,A1:A5&B1:B5,0)=ROW(A1:A5)-ROW(A1)+1))

or, if there could be blank cells in column A or B and you want to exclude
those, you could try:
=SUMPRODUCT((A1:A5<"")*(B1:B5<"")*(MATCH(A1:A5&B 1:B5,A1:A5&B1:B5,0)=ROW(A1:A5)-ROW(A1)+1))


" wrote:

Hi,

I am interested in finding a function that will count the number of
unique combinations a in two columns (columns A& B). For example:

Column A Column B Column C Column D
ABC 03/2007 =A1&B1 =SUMPRODUCT((1/
COUNTIF(C1:C5,C1:C5&"")))
BCA 03/2007 =A2&B2
CAB 07/2007 =A3&B3
ACB 07/2007 =A4&B4
ABC 03/2007 =A5&B5

In the case above, there are 4 unique combinations in columns A and B
(ABC 03/2007, BCA 03/2007, CAB 07/2007, ACB 07/2007). I don't count
the second instance of ABC 03/2007 becuase it is already in the first
row.

Is there a cleaner way of doing this? I've tried using sumproduct, but
I can't seem to get it to work the way I want it to. Does anyone know
how this can be done?

There are only four possible strings in column A (ABC, BCA, CAB, ACB)

Thanks!
Jason

Counting unique combinations in two columns
 

On May 3, 6:22 pm, JMB wrote:
You could also try
=SUMPRODUCT(--(MATCH(A1:A5&B1:B5,A1:A5&B1:B5,0)=ROW(A1:A5)-ROW(A1)+1))

or, if there could be blank cells in column A or B and you want to exclude
those, you could try:
=SUMPRODUCT((A1:A5<"")*(B1:B5<"")*(MATCH(A1:A5&B 1:B5,A1:A5&B1:B5,0)=ROW(A1:A5)-ROW(A1)+1))

" wrote:
Hi,

I am interested in finding a function that will count the number of
unique combinations a in two columns (columns A& B). For example:

Column A Column B Column C Column D
ABC 03/2007 =A1&B1 =SUMPRODUCT((1/
COUNTIF(C1:C5,C1:C5&"")))
BCA 03/2007 =A2&B2
CAB 07/2007 =A3&B3
ACB 07/2007 =A4&B4
ABC 03/2007 =A5&B5


great, thanks!

In the case above, there are 4 unique combinations in columns A and B
(ABC 03/2007, BCA 03/2007, CAB 07/2007, ACB 07/2007). I don't count
the second instance of ABC 03/2007 becuase it is already in the first
row.

Is there a cleaner way of doing this? I've tried using sumproduct, but
I can't seem to get it to work the way I want it to. Does anyone know
how this can be done?

There are only four possible strings in column A (ABC, BCA, CAB, ACB)

Thanks!
Jason