This must be a bug because its true for dividing by all odd numbers:
(i-1)/3
(i-2)/5
(i-3)/7
(i-4)/9
(i-5)/11
....
(i-n)/2n+1
And you can't make it work for even numbers
Added by Danny Boyes at 5:06pm on January 13, 2010
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…
branches in each A's list of B's, or remove its ends etcso that if I want to remove the last B in every A{0;1},{0;2},{0;3},{0;4},{0;5},{0;6}{1;1},{1;2},{1;3},{1;4}{2;1},{2;2},{2;3},{2;4},{2;5}would become{0;1},{0;2},{0;3},{0;4},{0;5}
{1;1},{1;2},{1;3}
{2;1},{2;2},{2;3},{2;4}I guess the question is do I need to figure out the cull pattern- each B may have different lengths...…
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
{1;1;4}{1;1;5}{1;2;0}{1;2;1}{1;2;2}{1;2;3}{1;2;4}{1;2;5}{1;3;0}{1;3;1}{1;3;2}{1;3;3}{1;3;4}{1;3;5}
etc...
and I want to format as a text it so it replaces the innermost branch with a letter so {0;0;1} would read A-0-1. I am able to replace all the symbols using replace text but am no sure if there's a way to convert a number to a letter.…
Added by Ryan Whitby at 12:40pm on February 3, 2015
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…
ve a Vertex [V] connected to four other Vertexs [N1-N4].
Each of the has a Value:
V ... 1
N1 ... 5
N2 ... 3
N3 ... 8
N4 ... 11
The Average Filter would set the Value of [V] to
(1+5+3+8+11)/5 = 5,6
The Median Filter would Sort Values and pick the middle one
1,3, [5], 8, 11
Hope that helped...…