4, 3, 7, 7, 7, 8
would need to be
5, 6, 7, 8, 4, 3, 7, 8
ideas?
current workaround (kangaroo dupPt component) removes all duplicates from list, tho does maintain order.
thx…
The best way is to use a C# or a VB component to transpose these
lists. I think in C# you can use transpose directly. You can ask this
on the VB/C# forum on our new website, www.grasshopper3d.com
- Scott
On May 27, 3:56 am, Tonsgaard wrote:
> Being a long time user of Generative Components trying to use
> grasshopper i miss the "transpose" command.
> I have a point list like this:
>
> 0, 1, 2, 3, 4, 5
> 0, 1, 2, 3, 4, 5
> 0, 1, 2, 3, 4, 5
> 0, 1, 2, 3, 4, 5
> 0, 1, 2, 3, 4, 5
>
> and a want to transpose dimensions to:
>
> 1, 1, 1, 1, 1
> 2, 2, 2, 2, 2
> 3, 3, 3, 3, 3
> 4, 4, 4, 4, 4
> 5, 5, 5, 5, 5
>
> Surely I am not the first in need of this...
> how would i go about and do this...? I suppose its quite easy in VB
> script, but being used to GC's C# like language, I kinda dont know how
> to do this...
>
> thanks...
>
> Tonsgaard…
points 0, X-1, (2*x)-1, (3*X)-1, (4*X)-1, (5*X)-1 and then
1, X, (2*x), (3*X), (4*X), (5*X)
2, X+1, (2*x)+1, (3*X)+1, (4*X)+1, (5*X)+1
and so on till
5, X+4, (2*x)+4, (3*X)+4, (4*X)+4, (5*X)+4
How can I do this best?
Thanks,
Niels…
the one-but-last list [4]. After running out of the n- items avalaible it should continue with the second item of list 0 and so on for all items on all the lists.
Intput, six lists of 30 items each
[0] (n=30)
[1] (n=30)
[2] (n=30)
[3] (n=30)
[4] (n=30)
[5] (n=30)
Output, 18 lists of 10 items each
[0],i=0;[5],i=4; [4],i=7;...
[0],i=1;[5],i=5; [4],i=8;...
...
[5],i=1;[4],i=5; [3],i=0;...
I thought perhaps the weave component or the relative tree item component but didn't manage to figure out how to compose the mask. I couldn't find much on how to use these. I guess it should wrap the lists, but not the items.
Any help would be greatly appreciated.…
Added by Thorsten Lang at 2:27am on January 24, 2011