;0;1;1;0}
{0;0;1;2;0} ...
{0;0;2;0;0}
{0;0;2;1;0}
{0;0;2;2;0} ...
{0;0;3;0;0}
{0;0;3;1;0}
{0;0;3;2;0} ...
...
and I would like to have this in two lists separated:
{0;0;0;0;0}
{0;0;0;1;0}
{0;0;0;2;0} ...
{0;0;2;0;0}
{0;0;2;1;0}
{0;0;2;2;0} ...
...
{0;0;1;0;0}
{0;0;1;1;0}
{0;0;1;2;0} ...
{0;0;3;0;0}
{0;0;3;1;0}
{0;0;3;2;0} ...
...
How can I do that?…
deos of this tutorial Series:
1. How to do CONCEPT DESIGN 3D Modelling in Rhino (Part 1 of 3); Beginner
https://www.youtube.com/watch?v=ZbMPZ...
2. How to do CONCEPT DESIGN: Material Textures in Rhino (Part 2 of 3)
https://www.youtube.com/watch?v=h8XRw...
3. How to do CONCEPT DESIGN -Create 3D Views in Rhino (Part 3 of 3);
https://www.youtube.com/watch?v=1RZMRwv8ub4
Feel Free to email us at Rhino4Arch@gmail.com for any help or information.…
.
Example input
Values : keys
Rectangle 1 : area 3
Rectangle 2 : area 1
Rectangle 3 : area 2
Output
Values : Keys
Rectangle 2 : area 1
Rectangle 3 : area 2
Rectangle 1 : area 3
So it seems you want to use area so use area of the surfaces into keys and keep the surfaces into values.…
Added by Michael Pryor at 2:03am on December 20, 2014
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…
t item (see the image), is it possible to do this in another way (quickly) ?
Is it possible to divide that curve into 2 separate curves using a point that i've used for the division?
Thanks…
Added by luca.pavarin at 4:08pm on January 7, 2010
e), {1;2}(line), {1;3}(line)... and on the other side to have {0;0}(all lines except {0}(0)), {0;1} (all lines except {0}(1)), {0;2}(all lines except {0}(2)), {0;3}(all lines except {0}(3)), {1;0} (all lines except {1}(0)), {1;1} (all lines except {1}(1)), {1;2} (all lines except {1}(2)) ,{1;3} (all lines except {1}(3))...The first tree is easy to achieve, simply grafting a branch for each element, and the other, what I've done is to copy all lines of each tree ({0},{1},{2},{3}), to have them in all branches of each tree ({0;0}(elements of {0}), {0;1}(elements of {0}),,{1;0}(elements of {1}), {1;1}(elements of {1})..., and then remove in the first branch({0;1} the first element(0), in the second branch the second element, the third branch the third element...And so correctly you compare each line with all the other within each branched tree.Aaaaapufff XD…