0, 5, 10, 15, 20
1, 6, 11, 16, 21
2, 7, 12, 17, 22
3, 8, 13, 18, 23
4, 9, 14, 19, 24
and if i'm here is because i'm not able... :)
can you help me?
thank you
…
nput parameter and then set the named values on the second?
protected override void BeforeSolveInstance() { Param_Integer param0 = Params.Input[0] as Param_Integer; Param_Integer param1 = Params.Input[1] as Param_Integer; param1.ClearNamedValues();
GH_Structure<GH_Integer> data = param0.VolatileData as GH_Structure<GH_Integer>; if (data.IsEmpty) return; foreach (GH_Integer value in data.AllData(true)) { switch (value.Value) { case 1: param1.AddNamedValue("First option for 1", 11); param1.AddNamedValue("Second option for 1", 12); param1.AddNamedValue("Third option for 1", 13); break;
case 2: param1.AddNamedValue("First option for 2", 21); param1.AddNamedValue("Second option for 2", 22); param1.AddNamedValue("Third option for 2", 23); break;
case 3: param1.AddNamedValue("First option for 3", 31); param1.AddNamedValue("Second option for 3", 32); param1.AddNamedValue("Third option for 3", 33); break; } return; } }
--
David Rutten
david@mcneel.com…
Added by David Rutten at 1:56am on December 18, 2013
created surface with sweep and rotate them with number of division. Section curves of 2Railsweep is defined with 3 Point arc and shape is also controlable.
You can change shape by sliding 16 paramteters.
The definition is rather long, hope somebody can modify with more sphiscated manner.
日本語
パラメトリックにボトルデザインをする.ghxファイルを作成しました。ボトルのプロファイルカーブを、参照となる円に対して、高さ、シフト、オフセットでコントロールポイントの位置情報を与えて定義。次にスイープで分割数だけ回転コピーして作成。
スイープの断面、3点円弧で作成し、形状コントロールするようにしています。
ボトル形状は、16のパラメーターをスライドすることで定義出来ます。
…
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
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
o now is select and group together list items by their index. {0:0} and {2:0} also {1:0} and {3:0}. They way I'm doing this at the moment is quite complicated and I'm sure there's an easier way to achieve the same result.
maybe someone in here knows a solution?
thanks a lot…
;1},{0;2},{0;3}, (note that the first item is NOT {0,0})
{1;1}{1;2},{1;3},
{2;1},{2;2},{2;3},
{3;1}{3;2},{3;3}...
Well I'll just upload an extract of the definition. The data path should be the same, but with the items in the last path shifted to the first path, and the items in the first path shifted to the second path and so on.
I'll keep trying. …