e! I do not have good ideas today!
The end result of the list would be:
5, 10, 15, 20, 21, (21 + 5), (21 + 10), (21 + 15), (21 + 20), (21 + 21), (42 + 5), (42 + 10), (42 + 15), (42 + 20), (42 + 21), etc …
output will show a tree with 3 branches of 4 integers each that I can pass on to other components. What is the best way to do it?
I have tried creating a tree and using a for loop to do so, but it didn't work.
Thank you for your help.
…
ee 3)
{5}
0 15
{6}
0 16
And I want to place points at every possible combination of these coordinates, treating Tree 1 as X coordinates, Tree 2 as Y coordinates, and Tree 3 as Z coordinates. Also, I would like the list of points to be a tree with paths corresponding to the coordinates. Wouldn't it be nice if I could plug these trees into a Point XYZ, with a new "branch cross reference" method, and get the following result?
{0:3:5}
0 {10.0, 13.0, 15.0}
{0:3:6}
0 {10.0, 13.0, 16.0}
{0:4:5}
0 {10.0, 14.0, 15.0}
{0:4:6}
0 {10.0, 14.0, 16.0}
{1:3:5}
0 {11.0, 13.0, 15.0}
{1:3:6}
0 {11.0, 13.0, 16.0}
{1:4:5}
0 {11.0, 14.0, 15.0}
{1:4:6}
0 {11.0, 14.0, 16.0}
{2:3:5}
0 {12.0, 13.0, 15.0}
{2:3:6}
0 {12.0, 13.0, 16.0}
{2:4:5}
0 {12.0, 14.0, 15.0}
{2:4:6}
0 {12.0, 14.0, 16.0}
In this form of cross referencing, every combination of individual branches from the different lists is used as separate input, and the output for each combination is put onto a branch in the result whose path is the concatenation of the input branch paths used.…
Added by Andy Edwards at 7:03pm on November 3, 2009
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…
First use a series component with start=1, step=42, count=3
Use the output to create a new series component with start=existingseries, step=1, count=11
rve
10 curve
11 curve
12 curve
13 curve
...and I'd like to rearrange the order in which the curve are listed, to something like this:
{0,0,0}
0 curve
1 curve
8 curve
9 curve
10 curve
11 curve
2 curve
3 curve
4 curve
5 curve
12 curve
13 curve
6 curve
7 curve
I hope this makes sense.
Thank in advance for any advice,
John…