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…
e), {1;2}(line), {1;3}(line)... and on the other side to have {0;0}(all lines except {0}(0)), {0;1} (all lines except {0}(1)), {0;2}(all lines except {0}(2)), {0;3}(all lines except {0}(3)), {1;0} (all lines except {1}(0)), {1;1} (all lines except {1}(1)), {1;2} (all lines except {1}(2)) ,{1;3} (all lines except {1}(3))...The first tree is easy to achieve, simply grafting a branch for each element, and the other, what I've done is to copy all lines of each tree ({0},{1},{2},{3}), to have them in all branches of each tree ({0;0}(elements of {0}), {0;1}(elements of {0}),,{1;0}(elements of {1}), {1;1}(elements of {1})..., and then remove in the first branch({0;1} the first element(0), in the second branch the second element, the third branch the third element...And so correctly you compare each line with all the other within each branched tree.Aaaaapufff XD…
W, X, Y, Z}
----------------------------------------
and if I set this
SetA = {U, V, W, X, Y, Z}
SetB = {1, 2, 3}
Imap = {2, 2, 2}
I will get this?
result = {U, V, 3, 2, 1, W, X, Y, Z}
----------------------------------
And what if I set this?:
SetA = {U, V, W, X, Y, Z}
SetB = {1, 2, 3}
Imap = {2, 2}…
Added by Frane Zilic at 3:26pm on September 10, 2010
ctivity of vertices ordered sequentially, the order defining the direction of the normal, using 0 1 3 2 Causes an error this way. If it a quad face it seems odd to me that you would label the vertices in such an order, as an engineer, i have never seen it done as 0 1 3 2, it could be 3 2 1 0, 2 1 0 3 etc but going 0 1 3 2 is not acceptable, i will do a bit more reading on this.…
Added by Steve Lewis at 5:18pm on December 24, 2013