is shorthand for [0 to 8].
> 10 Any number larger than X. This notation is shorthand for [11 to infinity].
>= 5 Any number larger than or equal to X. This notation is shorthand for [5 to infinity].
--
David Rutten
david@mcneel.com
Seattle, WA…
Added by David Rutten at 9:27pm on November 3, 2013
se the cull pattern, so I wanted to make the pattern using a function component. x=y. x= the original list and y= the interval i wanted to remove. So the pattern should be:
0: false
1:false
2:false
3:false
4:true
5:true
6:true
7:true
8:false
9:false
10:false
etc...…
Added by Rasmus Holst at 3:32am on November 17, 2009
uld be much better than Rhino at huge mesh collections. I'd personally try free Autodesk Meshmixer and ZBrush first but most designers are more familiar with rendering programs like Maya or 3DS Max. I'm not familiar enough with architecture to suggest a list as only Revit and Sketchup come to mind.
Looking more closely, CAD Exporter is only for 2D curves and points, how silly, and it requires baked geometry in a Rhino layer:
I could write a Python script to export an STL but that would be a large ascii format file instead of binary. Better to use OBJ to retain quad faces, too.
Ah, well, OBJ files are also ascii format when exported from Rhino, so it would be quite easy to make a script to export those directly to disk from Grasshopper. Here is one box, 10X10X20 in size, with quad faces:
# Rhino
o object_1v 10 10 20v 10 10 0v 10 0 20v 10 0 0v 0 10 20v 0 10 0v 0 0 20v 0 0 0f 5 7 3 1f 5 6 8 7f 3 7 8 4f 2 4 8 6f 5 1 2 6f 3 4 2 1
If I have time I'll make a little script to write such OBJ files unless you can find a native Grasshopper plugin for direct OBJ export in full 3D for meshes.…
a follow up question... how do I wrap a list onto itself at a certain frequency?
i.e. I want the list {1;2;3;4;5;6;7;8;9}
to become {1,4,7; 2,6,8; 3,6,9} wrapped every 3rd item
Added by Joshua Jordan at 5:30pm on November 17, 2012
opper is all these values "recognizing" as similar/same.
I got list of results (n) with following values:
0. -3.2584e-9 1. -4.4992e-9 2. -6.7220e-9 3. -4.5154e-9 4. -4.3325e-9 5. -2.2496e-9 6. -2.2385e-9 7. -6.7525e-9 8. -4.5154e-9
Even though most of these values (maybe all of them) "go" into the second group:
(10^(-9)≤n) and (n>10^(-4))
Grasshopper recognizes all of them as members of the first group:
10^(-4)≥n
I am aware that this kind of very small values are unusual, and maybe Grasshopper is not made for it. But is there any way this can be done?
Take a look:
Thank you.…
Integer = 0 To 9
val *= 2
lst.Add(val)
Next
Since val is a ValueType, when we assign it to the list we actually put a copy of val into the list. Thus, the list contains the following memory layout:
[0] = 2
[1] = 4
[2] = 8
[3] = 16
[4] = 32
[5] = 64
[6] = 128
[7] = 256
[8] = 512
[9] = 1024
Now let's assume we do the same, but with OnLines:
Dim ln As New OnLine(A, B)
Dim lst As New List(Of OnLine)
For i As Integer = 0 To 9
ln.Transform(xform)
lst.Add(ln)
Next
When we declare ln on line 1, it is assigned an address in memory, say "24 Bell Ave." Then we modify that one line over and over, and keep on adding the same address to lst. Thus, the memory layout of lst is now:
[0] = "24 Bell Ave."
[1] = "24 Bell Ave."
[2] = "24 Bell Ave."
[3] = "24 Bell Ave."
[4] = "24 Bell Ave."
[5] = "24 Bell Ave."
[6] = "24 Bell Ave."
[7] = "24 Bell Ave."
[8] = "24 Bell Ave."
[9] = "24 Bell Ave."
To do this properly, we need to create a unique line for every element in lst:
Dim lst As New List(Of OnLine)
For i As Integer = 0 To 9
Dim ln As New OnLine(A, B)
ln.Transform(xform)
lst.Add(ln)
Next
Now, ln is constructed not just once, but whenever the loop runs. And every time it is constructed, a new piece of memory is reserved for it and a new address is created. So now the list memory layout is:
[0] = "24 Bell Ave."
[1] = "12 Pike St."
[2] = "377 The Pines"
[3] = "3670 Woodland Park Ave."
[4] = "99 Zoo Ln."
[5] = "13a District Rd."
[6] = "2 Penny Lane"
[7] = "10 Broadway"
[8] = "225 Franklin Ave."
[9] = "420 Paper St."
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 6:26am on September 9, 2010
lues. What I want to do is combine them so that the structure would be something like:
{4;0}
{4;1}
{4;2}
{4;3}
{5;0}
{5;1}
{5;2}
{5;3}
I tried the method here, but it didn't give me what I wanted, it was just tacking the new values onto the end, and not maintaining their paths. Any help would be appreciated. Thanks!…
Added by Dennis Goff at 8:13am on February 10, 2016