etc.
Group 2 - 1, 6, 11, 16, 21 etc.
Group 3 - 2, 7, 12, 17, 22 etc.
Group 4 - 3, 8, 13, 18, 23 etc.
Group 5 - 4, 9, 14, 19, 24 etc. "
except in data, the branches start at 0, so 'group 1' is branch 0
as for the order of your points, that depends on the input prior sorting...
yrs …
{4}-0;3
{5}-6;7
{6}-5;7
{7}-5;6
Here it can be shown that there are two subgraphs containing 0,1,2,3,4 and 5,6,7. How can I use spiderweb (either using scripting or the components) to give me this result when I have many more vertices??
Thanks,
Sam…
Maybe with numbers it would be easier to understand.
If there is 1 insert to the list 5
If there is a 5 insert to the list 6 or 4
So if i start from 1 one possible list would be: {1, 5, 6}
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…