Unique Values

aabbasi

Board Regular
Joined
Mar 4, 2002
Messages
188
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.
 

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lenze

Legend
Joined
Feb 18, 2002
Messages
13,690
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

Well-known Member
Joined
Oct 16, 2002
Messages
657
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

Board Regular
Joined
Mar 4, 2002
Messages
188
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

Well-known Member
Joined
May 20, 2002
Messages
890

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Using the add-in try

COUNTDIFF(RANGE)
 

aabbasi

Board Regular
Joined
Mar 4, 2002
Messages
188

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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

Well-known Member
Joined
Feb 15, 2002
Messages
4,967
Office Version
  1. 365
  2. 2010
Platform
  1. Windows
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

MrExcel MVP
Joined
Feb 14, 2002
Messages
85,209
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

Board Regular
Joined
Mar 4, 2002
Messages
188
Thanks Aladin...
I guess I am not Smart enough to follow your good formula. I gave up!
 

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