a specific domain, for example:
0.) 0 to 1 -----> 11 random values from 0 to 1 (0.245,0.678,0.36,0.78,.28,0.18........)
1.) 1 to 2 -----> 11 random values from 1 to 2 (1.26,1.36,1.01,1.68,1.26,1.96.........)
3.) 2 to 3 -----> 11 random values from 2 to 3 (2.96,2.45,2.78,2.56,2.98,2.10..........)
4.) 3 to 4 and so on where I have a data set containing 11 paths with 11 values and the values fall within the specific domain.
Like my post above I have the correct path but I need to feed it the correct seed to get different values for each number. I tried grafting a series similar to the last post but it scrambles my data. Thanks so much for the help!
…
nts me this:
[[0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7]]
this is what I wanted but how to convert this to tree in grasshopper?
In grasshopper I just get:
8x IronPython.Runtime.List…
{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
dont get you, i am saying sleect numbers in range 1 to 10, starting from 1 with a step of 2.
1 to 10 by 3 = 1 4 7 10
1 to 10 by 5 = 1 6
1 to 10 by 1 = 1 to 10 = 1 2 3 4 5 6 7 8 9 10
Added by Steve Lewis at 3:15pm on November 11, 2013
the curves on surface issue it's solved seting flatten to the surface control point output. Still didnt know how to group points like:
1;1, 2;2, 3;3.....
1;2, 2;3, 3;4....
1;3, 2;4, 3;5...
....
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
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...…
} (N=11) {0;1} (N=11) {0;2}(N = 11) {0;3}(N = 11) {0;4}(N = 11)
2. I run the Points that are coming out from the Divide Curve Components through the Path Mapper components with this definition:
{A;B} (i) > {A} (i)
3. I run data coming out from Path Mapper component through:
a) Parameter Viewer component and the result is:
{0} N=11 (data with 1 branches)
b) Point > Panel and the result is:
collection of 11 point (N=11) which is the exactly the same as the collection of point belonging to {0;4} (N = 11).
So, here is the question:
why the collection of points coming out from the Path Mapper {A;B} (i) > {A} (i) component is the same as the collection of points belonging to the curve {0;4}(N = 11) ?
Anyway ... It 's the first time I ask a question here... so I would like to thank you for what you do with your work! Thank you! You are really great!…