0;3} (N = 2)
{0;0;0;4} (N = 2)
{0;0;1;0} (N = 2)
{0;0;1;1} (N = 2)
{0;0;1;2} (N = 2)
{0;0;1;3} (N = 2)
{0;0;1;4} (N = 2)
Flattening this structure using the Flatten component would result in:
{0} (N = 20)
However, using a Path Mapper with the following masks will flatten is somewhat more intelligently:
{A;B;C;D} -> {A;B;C}
Now, you get:
{0;0;0} (N = 10)
{0;0;1} (N = 10)
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 3:19am on December 14, 2009
tion and the points array that follow. I was used to the Rhino.EvaluateSurface(Strobj, ArrParam) to pick pints on a surface.
Then I understand that points are identified with var pt and added with function pt_list.Add(pt) . But can't understand the whole syntax. This code is supposed to create hexagons!
Dim lines As New List(Of OnPolyline)()
'Point Culling
Dim ptArr As New On3dPointArray()
Dim pt_list As New List(Of On3dPoint)
For i As Int32 = 0 To u - 2 Step 2
For j As Int32 = 0 To v - 3 Step 4
ptArr.Append(arrPt(i + 1, j).x, arrPt(i + 1, j).y, arrPt(i + 1, j).z)
ptArr.Append(arrPt(i + 2, j + 1).x, arrPt(i + 2, j + 1).y, arrPt(i + 2, j + 1).z)
ptArr.Append(arrPt(i + 2, j + 2).x, arrPt(i + 2, j + 2).y, arrPt(i + 2, j + 2).z)
ptArr.Append(arrPt(i + 1, j + 3).x, arrPt(i + 1, j + 3).y, arrPt(i + 1, j + 3).z)
ptArr.Append(arrPt(i, j + 2).x, arrPt(i, j + 2).y, arrPt(i, j + 2).z)
ptArr.Append(arrPt(i, j + 1).x, arrPt(i, j + 1).y, arrPt(i, j + 1).z)
ptArr.Append(arrPt(i + 1, j).x, arrPt(i + 1, j).y, arrPt(i + 1, j).z)
Dim pline As New OnPolyline(ptArr)
lines.Add(pline)
Dim pt As New On3dPoint((arrPt(i + 1, j).x + arrPt(i + 1, j + 3).x) / 2, (arrPt(i + 1, j).y + arrPt(i + 1, j + 3).y) / 2, (arrPt(i + 1, j).z + arrPt(i + 1, j + 3).z) / 2)
pt_list.Add(pt)
ptArr.destroy
Next
Next…
Added by Jon Malkovich at 7:16pm on November 3, 2009
6, 7, 8, 9, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, etc
In addition to the components in the attached file, I have also tried using Cull Index but that did not do much. I tried using a number slider set to whole numbers with the range equal to the values I have set up in the integer, but it just increases the size by 1 every time. Any help on steps in the right direction would be great thanks.…
;-2;-1;0;1;2;3) and beside that i don't want to have 0 as a result.
So at the End I would like to have only 6 numbers: -3;-2;-1;1;2;3
I already came to an result with integer numbers but it actually only rounded up the random results. when one of the results was -0.473 it ended up as 0.
I'm sure it's pretty easy to solve, I'm just missing something.
thx for help…