i=1 -- array2(1) = 3 (occurrences of "2"),
i=2 -- array2(2) = 2 (occurrences of "4"),
i=3 -- array2(3) = 2 (occurrences of "3").

Hi. One idea is to use a Dictionary Object. Here is a general idea.
I understand you only want an array of those items with a count 1.

Sub Demo_Tally()
Dim d
Dim p
Dim n, k
Dim v
v = Array(1, 2, 4, 6, 5, 4, 7, 2, 3, 2, 3)

Set d = CreateObject("Scripting.Dictionary")
' Add Key, Item (Both Required)

'// Tally items
For p = LBound(v) To UBound(v)
n = v(p)
If d.exists(n) Then
d(n) = d(n) + 1
Else
d.Add n, 1
End If
Next p

'// Remove items with Count = 1
For Each k In d.keys
If d(k) = 1 Then d.Remove (k)
Next k

'// Display items
For Each k In d.keys
Debug.Print k, d(k)
Next k
End Sub


Returns:
Item | Count
2 3
4 2
3 2

--
HTH :)
Dana DeLouis
Windows XP & Excel 2007


"a.riva@UCL" wrote in message
ups.com...
Thanks for all the suggestions!

The code that Dana sent is working very well, and it's really
simple :-)

Now I have an other question...

I have my usual option-based-1 array1 in VBA, which contains x
numbers. Some of them are repeated. What I would like to do is
creating an other option-based-1 array, let's call it array2, which
contains the numbers of occurrences of each of the repeated elements
of array1 within array1... I'm struggling to find a solution...

For example:

option-based-1 array1 is (1, 2, 4, 6, 5, 4, 7, 2, 3, 2, 3) -- I
cannot sort the array.

I think that the procedure should do the following operation: it
detects how many items are repeated in "array1", and for each of this
repeated items stores in a new array "array2" a number corresponding
to the number of its occurrences.

For example, in array1 the procedure detects that there are n=3 items
which occur more than once (they are "2", "4" and "3"). Then it ReDims
array2 (1 to n), and for i=1 to n it gives to array2(i) the values of:

i=1 -- array2(1) = 3 (occurrences of "2"),
i=2 -- array2(2) = 2 (occurrences of "4"),
i=3 -- array2(3) = 2 (occurrences of "3").

Can somebody help me?

Thanks in advance :-)

Antonio.