I have this :
list 3 : 0 1 2 3 4 5 6
list 2 : 0 1 2 3 4 5 6
list 1 : 0 1 2 3 4 5 6
list 0 : 0 1 2 3 4 5 6
and I want to group the points of index 0 in a branch, the points of index 1 in another branch and so on.
I attached a file in which I generated the points.
Thank you in advance for your help !
Regards
Red…
Your original tree contains 3 items, your desired three contains 9 items. Where do these 6 new items come from?
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David Rutten
david@mcneel.com
Poprad, Slovakia
e possible to change the component definition making possible to customize the number of outputs.Now Dispatch moves "true" values to A and "False" values to B
INPUT:
L (List to work on) -> 1, 2, 3, 4, 5, 6, 7, 8
D (Dispatch Pattern) -> True, False
OUTPUT:
A (List) -> 1, 3, 5, 7
B (List) -> 2, 4, 6, 8
Could it be possible/useful to modify it so it could dispatch items to several outputs, like:
INPUT:
L (List to work on) -> 1, 2, 3, 4, 5, 6, 7, 8, 9, 0
D (Dispatch Pattern) -> A, B, C
OUTPUT:
A (List) -> 1, 4, 7, 0
B (List) -> 2, 5, 8
C (List) -> 3, 6, 9
maybe I'm missing something and there's already a component with this function... I have been searching on the forum for half afternoon, but can't find anything about it!
Thank you!…
ll these 12500 points.
Group 1 would represent the point located at 0, 5, 10, 15, 20 etc.
Group 2 - 1, 6, 11, 16, 21 etc.
Group 3 - 2, 7, 12, 17, 22 etc.
Group 4 - 3, 8, 13, 18, 23 etc.
Group 5 - 4, 9, 14, 19, 24 etc.
I can create the pattern but the selection of points are all the points in row 0 and then all the points in row 5 and so on.
I would like the selection of points to start at the bottom left, and sequentially continue to the right and then continue on the 2nd row (left to right & bottom to top). i am hoping the pattern i am trying to achieve is more understood with the quick screen capture I uploaded.
the end goal is to be able to select all the points in the grid that are in each pattern.
Thanks in advance for any guidance with this. …
Added by Alyne Rankin at 6:53am on October 11, 2017
{0;1;0}N=6
{0;1;1}N=6
{0;1;2}N=5
{0;2;0}N=7
{0;2;1}N=8
{0;2;2}N=9
Can you shift and wrap any of the paths A B or C?
Say if I wanted to shift and wrap B by 1 to get the following...
{0;0;0}N=7
{0;0;1}N=8
{0;0;2}N=9
{0;1;0}N=3
{0;1;1}N=2
{0;1;2}N=5
{0;2;0}N=6
{0;2;1}N=6
{0;2;2}N=5…
could accomplish this by adding something like '5.0 + Slider * 3.0' into the expression field... (and having an integer slider with domain 0 to 6).
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 10:19am on February 24, 2011