Unique Values

[aabbasi]

I have added the following addin: MoreFunc from
http://longre.free.fr/english/index.html,

There is a function called {=UNIQUEVALUES(Array,Order)}. My problem is that it is not giving the list of Unique values instead it is showing only one Value. Does anyone has experience working with this in Excel 2000.

Thank you.

Eg: Values 23,23,45,34,32 should show 45,34,32,23 but instead of this it is showing 45 only.

 

[lenze]

Say your array is in A2:A10. Select B2:B10 and enter the formula =UNIQUEVALUES(A2:A10,0)
Becasue this is an array formul, you must enter it using Ctrl+Shift+Enter

 

[Chas17]

Seems like it is working to me, 45 is unique, the digits 4 and 5 are unique to that number, all other numbers in the series have a common, 23 and 32 have the digit 2; 34 and 32 have the digit 3....

 

[aabbasi]

Lenze & Charles17:

Thank you for your quick reaponse.

What I am trying to achieve is that lets say if I have a list of numbers (Initial List of 8 numbers):
12, 23, 34, 44, 55, 55, 65, 23
I am trying to run the function to get unique values (6 numbers) as : 12, 23, 34, 44, 55, 65 which means the selection from initial list minus repeated numbers (in this case 55 & 23)

 

[maxflia10]

Using the add-in try

COUNTDIFF(RANGE)

 

[Aladin Akyurek]

See

http://www.mrexcel.com/board/viewtopic.php?topic=8659&forum=2

for how to combine INDEX with UNIQUEVALUES to get the desired result.

This function creates/computes an array constant. When the computation occurs in a single cell, you get to see the leftmost element of the computed array constant, completely in accord with Excel's own specs.

 

[aabbasi]

Maxflia: Thank you.
Using CountDiff gives me the COUNT how many are unique, but I want to see actual data which is unique.

Aladin Thank you...but its kind of lengthy process.

 

[Dave Patton]

It is not a lengthy process.

With your numbers in A1:A20

put the following in B1 and copy or FillDn as
far as required.

=INDEX(UNIQUEVALUES($A$1:$A$20,1),ROW(1:1))
This message was edited by Dave Patton on 2002-10-30 11:57

 

[Aladin Akyurek]

On 2002-10-30 11:45, aabbasi wrote:
Maxflia: Thank you.
Using CountDiff gives me the COUNT how many are unique, but I want to see actual data which is unique.

Aladin Thank you...but its kind of lengthy process.

Try:

=LEFT(SETV(MCONCAT(UNIQUEVALUES(Range,1),",")),LEN(GETV())-1)

 

[aabbasi]

Thanks Aladin...
I guess I am not Smart enough to follow your good formula. I gave up!