Count distinct function?...

lenny3k

New Member
Joined
Jan 15, 2003
Messages
18
Hi there.

Is there a way to count distinct values in a range of cells in a worksheet?

i'm looking for something like a COUNT DISTINCT function that would count all distinct values in an entire column. any ideas?

any help would be appreciated!

Len
 

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fairwinds

MrExcel MVP
Joined
May 15, 2003
Messages
8,638
Du you mean count unique?
(English is not my native tongue)


Then try:

=SUMPRODUCT((A1:A10<>"")/COUNTIF(A1:A10,A1:A10&""))
 

luckycharm

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Joined
Sep 3, 2003
Messages
88
Fairwinds,
Care to elaborate on the qriteria used in the countif arguments?
It does work, but not sure about the intuition.
Cheers.
 

fairwinds

MrExcel MVP
Joined
May 15, 2003
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8,638
Hi,
I learned this formula from Aladin but I cannot find a post to refer to where he explains it.

Anyway, The first part gives 1 for each cell that is not empty. Then divides it with how many instances there is of this value.
Summing that gives you the "unique count".

The &"" is to avoid 0 and #DIV/0 when a cell is empty.
 

luckycharm

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Joined
Sep 3, 2003
Messages
88
So (as usual) everything begins with Aladin.
Sould've guessed. :whistle:

Hopefully he reads in on this one and elaborates, since I
cannot say I follow/agree with your explanation.
Anyhow, to me it looks as though the second part (countif)
returns number of occurences for the "non distinct".

Waiting on Aladin.
 

Aladin Akyurek

MrExcel MVP
Joined
Feb 14, 2002
Messages
85,170
luckycharm said:
So (as usual) everything begins with Aladin.
Sould've guessed. :whistle:

Hopefully he reads in on this one and elaborates, since I
cannot say I follow/agree with your explanation.
Anyhow, to me it looks as though the second part (countif)
returns number of occurences for the "non distinct".

Waiting on Aladin.
I did have two Invalid_Session in trying to respond to this one.

Well, the formula Fairwinds posted is due to Harlan Grove, although I did play the role of an evaluator regarding the earlier public versions Harlan posted at the worksheet.functions.

Harlan eventually succeeded to create a SumProduct version of David Hager's formula, which is:

{=SUM(1/COUNTIF(Range,Range))}

The kernel idea of this formula is:

1/Tokens(Type)

Tha is, if Bob is a distinct name (a type) and Bob occurs say 3 times, we can assign a weight to each of its occurrences (tokens), thus:

1/3, 1/3, 1/3

Summing the individual weights we get 1 back. That is: dividing 1 by the tokens, we get a distinct/unique/type count.

Two things to note about this formula:

If Range has any empty cell, we get #DIV/0! for COUNTIF(EmptyCell,EmptyCell) == 0.

If Range has any cell housing a formula (like =IF(X1,1,"") that returns a blank (sometimes referred to as null string), COUNTIF(CellWithFormulaBlank,CellWithFormulaBlank), we get a count of 1 for such a cell.

It's somewhat theoretical whether an empty cell should be considered a distinct type. The same holds for a formula-blank. Supposing that they are not distinct types, Hager's formula needs some modification before it can be applied to a Range housing empty cells or formula blanks:

{=SUM(IF(LEN(Range),1/COUNTIF(Range,Range))}

or

{=SUM(IF(Range<>"",1/COUNTIF(Range,Range))}

It's obvious that using the latter (with lesser function call) is better (for robustness).

My contribution (along with Juan) consists of just this modification.

How does this formula works using an example?

Let A2:A8 house the following sample:

{"Bob";"Bob";"Bob";"Jane";"Jane";EmptyCell;"Thomas"}

where EmptyCell stands for an empty cell (not for a distinct type).

The COUNTIF(A2:A8,A2:A8) bit gives:

{3;3;3;2;2;0;1}

The 1/COUNTIF(A2:A8,A2:A8) gives:

{0.333333333333333;0.333333333333333;0.333333333333333;0.5;0.5;#DIV/0!;1}

Note #DIV/0!. Clearly, there are 3 types/distinct/unique items if SUM could ignore the error values.

With

{=SUM(IF(A2:A8<>"",1/COUNTIF(A2:A8,A2:A8)))}

we have successively:

=SUM(IF({TRUE;TRUE;TRUE;TRUE;TRUE;FALSE;TRUE},{0.333333333333333;0.333333333333333;0.333333333333333;0.5;0.5;#DIV/0!;1}))

When IF does effect the filtering:

=SUM({0.333333333333333;0.333333333333333;0.333333333333333;0.5;0.5;FALSE;1})

Since SUM ignores logical values, we get: 3.

It is quite easy to turn Hager's original formula

{=SUM(1/COUNTIF(Range,Range))}

into a SumProduct formula:

=SUMPRODUCT(1/COUNTIF(Range,Range))

but it isn't regarding:

{=SUM(IF(Range<>"",1/COUNTIF(Range,Range)))}

Whenever IF is needed when computing with array objects, a control+shift+entered formula is almost always a necessity.

Harlan eventually arrived at:

=SUMPRODUCT((A2:A8<>"")/COUNTIF(A2:A8,A2:A8&""))

which is harder to understand, but becomes intelligible if one knows the following about CountIf (as touched upon at the beginning of this post)...
UitlegHagerGrove.xls
ABCD
1
2
0
3
 
1
4
0
5
1
6
Sheet3


C2:

=COUNTIF(A2,A2)

C5:

=COUNTIF(A5,A5&"")

Using our original example...

The (A2/A8<>"") bit gives:

{TRUE;TRUE;TRUE;TRUE;TRUE;FALSE;TRUE}

The COUNTIF(A2:A8,A2:A8&"") bit gives [ see the foregoing about the behavior of CountIf regarding "" ]...

{3;3;3;2;2;1;1}

The (A2:A8<>"")/COUNTIF(A2:A8,A2:A8&"") bit gives:

{TRUE;TRUE;TRUE;TRUE;TRUE;FALSE;TRUE}/{3;3;3;2;2;1;1}

The pairwise division (with coercion of logical values into numbers) gives:

{0.333333333333333;0.333333333333333;0.333333333333333;0.5;0.5;0;1}

which gets summed, producing as result: 3.

Concatenating Range with "" recurs also in other forms in other formulas where filtering If can be circumvented like in:

http://www.mrexcel.com/board2/viewtopic.php?t=70140&highlight=left
 

luckycharm

Board Regular
Joined
Sep 3, 2003
Messages
88
So this is what a concise dictionary looks like.
Reading is weeping :cry:
Sorry for your missed sessions.
Took me 3 to grasp :LOL:

Cheers.
 

marrydavidson101

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I recently done my master in Business Administration and now days I want to do some online Microsoft certifications like MCSA MCSE Oracle java etc and I Found this online notes is a best helping source to get online accurate information which is more helpful for my preparation.
 

daveWP

New Member
Joined
Sep 29, 2009
Messages
28
Very interesting discussion.

I have another issue that adds another dimension to this problem though.

What if I had a data range that contained numerous child records.

And lets say I needed to count the total number of distinct records for each Master record.

Here's a sample data range:

A B C
1 1 Accounting
1 1 Accounting
1 1 Supply
1 2 Accounting
1 2 Marketing
2 1 Supply
2 1 Accounting
2 1 Supply
2 2 Supply

the distinct record count is a combination of A, B, C.

Assume on another sheet, we have A1 = 1, A2 = 2
B1 is array formula that wil give us the answer 4, and B2 yields 3

Any ideas?
 

Domenic

MrExcel MVP
Joined
Mar 10, 2004
Messages
19,139
For simplicity, let's assume that A2:C10 contains the data, E2 contains 1, and E3 contains 2. Then, assuming that the data does not contain empty/blank cells, try...

F2, confirmed with CONTROL+SHIFT+ENTER, and copied to down to F3:

=SUM(IF(FREQUENCY(IF($A$2:$A$10=E2,MATCH("~"&$A$2:$A$10&$B$2:$B$10&$C$2:$C$10,$A$2:$A$10&$B$2:$B$10&$C$2:$C$10&"",0)),ROW($A$2:$A$10)-ROW($A$2)+1)>0,1))
 

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