I can only think of using the script component for this.
Try the following code:
Dim bol(x.count-1) As Boolean
For i As Integer = 0 To x.count - 1
bol(i) = True
For j As Integer = i + 1 To x.count - 1
If x(i) = x(j) Then bol(i) = False
Next
Next
a = bol
In the "x" input of the scripting component you input the list of
numbers (set it as 'list').
It will output the "true, false, ..." pattern that should be linked to
the cull pattern component.
This code was improvised by me, i have very basic notions of
programming, I'm not sure if it works properly, try it first. there's
probably a better way to do it.
You could probably output directly the culled list of numbers, but i
think that it's better to do as much as possible outside the scripting
component. (at least it's more fun).
On Nov 26, 5:35 pm, Rchitekt <[EMAIL PROTECTED]> wrote:
> Thanks visose,
> That makes sense. I knew there must have been an easy way to do it.
> My only problem is that there could be cases where there are more than
> one duplicate. So for the case above, there could be three 4's, or
> more. The list should already be ordered, so they should always be
> next to each other. But is there a way to shift more than 1 number,
> without using multiple shift and cull components?
> -Andy
>
> On Nov 26, 12:21 am, visose <[EMAIL PROTECTED]> wrote:
>
> > If there is at most only 1 duplicate of a number, and you have one
> > next to the other (as it is on your example) or you don't mind re-
> > ordering the list so that they follow each other, you can use the
> > shift component with the shift offset input as 1. Then connect the
> > original list and the offseted list to a 2 input formula with the
> > following expression: x≠y.
> > Then connect the original list and the formula component to a cull by
> > pattern component.
> > What the expression does is that for every item of the original list
> > it checks if the next item is a different number. If it is, it returns
> > true.
>
> > On Nov 26, 8:43 am, Rchitekt <[EMAIL PROTECTED]> wrote:
>
> > > I know there is a very simple way to do this, but for some reason, the
> > > method is escaping me. Let's say I have a list of 10 numbers:
> > > 1,2,3,4,4,5,5,6,7,8. So, there are two 4's and two 5's. What I would
> > > like to do is search the list and if there is a duplicate number, then
> > > report a true value for the first one, and a false value for the
> > > duplicate. Ultimately, I would just like my list that originally had
> > > 10 numbers to have 8, which would look like 1,2,3,4,5,6,7,8. I could
> > > then feed that boolean list to a Dispatch component to split the
> > > list. Is there any easy way to do this. Maybe it's just late and my
> > > brain isn't working properly. Thanks for the help.
> > > -Andy
