Count distinct function?...

[lenny3k]

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

[fairwinds]

Du you mean count unique?
(English is not my native tongue)


Then try:

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

[luckycharm]

Fairwinds,
Care to elaborate on the qriteria used in the countif arguments?
It does work, but not sure about the intuition.
Cheers.

[fairwinds]

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]

So (as usual) everything begins with Aladin.
Sould've guessed.

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]

Originally Posted by luckycharm

So (as usual) everything begins with Aladin.
Sould've guessed.

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)...

******** ******************** ************************************************************************>

Microsoft Excel - UitlegHagerGrove.xls ___Running: xl2000 : OS = Windows Windows 2000
(F)ile (E)dit (V)iew (I)nsert (O)ptions (T)ools (D)ata (W)indow (H)elp (A)bout
B2A3B3B4B5 =
	A	B	C	D
1	*	*	*	*
2	*	0	*	*
3	*	1	*	*
4	*	0	*	*
5	*	1	*	*
6	*	*	*	*
Sheet3 *

[HtmlMaker 2.32] To see the formula in the cells just click on the cells hyperlink or click the Name box
PLEASE DO NOT QUOTE THIS TABLE IMAGE ON SAME PAGE! OTHEWISE, ERROR OF JavaScript OCCUR.

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/viewto...highlight=left

[luckycharm]

So this is what a concise dictionary looks like.
Reading is weeping
Sorry for your missed sessions.
Took me 3 to grasp

Cheers.