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
0;5} n8 I can easily retrieve a specific tree branch, then modify the data, lets say to each have 6 items in their list instead of 8. the end resut i want is:
{0;0} n6
{0;1} n8
{0;2} n8
{0;3} n8
{0;4} n8
{0;5} n6
But i cant seem to get the "replace branch" componet to work
In reality i have a tree that consists of 108 branches, so exploding the tree and entwining (which is my normal work around) is impractical
any help is appreciated!…
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!…
order away. After been split the surfaces don't have the same order anymore.
What I need: ...in branch (0;0) is the surface between curves 0;0 / 0;0 / 0;1...in branch (0;1) is the surface between curves 0;1 / 0;2...in branch (0;2) is the surface between curves 0;2 / 0;3...in branch (0;3) is the surface between curves 0;3 / 0;4...in branch (0;4) is the surface between curves 0;4 and outer boundary...in branch (0;5) all other surfaces
But i can not do it manually with list item because this is just a test, sometimes I have much more curves and the partition before the splitting needs to give the clue, how the surfaces are branched after the splitting.
Does any have an idea how to fix that? I could also be done without splitting, just with the curves, but I don't know how, if everthing needs to stay parametricThanks in advance…
ap value = True
Shift List = 1 --> (B,C,D,A)
Shift List = 2 --> (C,D,A,B)
You can also use negative values.
Shift List = -1 --> (A,B,C)
Shift List = -2 --> (A,B)
and with Wrap = True
Shift List = -1 --> (D,A,B,C)
Shift List = -2 --> (C,D,A,B)
The most useful Shift List action I use is to either get rid of the first or last item in a list and sometimes both.
Shift list = -1 --> (A,B,C) Shift list = 1 --> (B,C)
In the example posted above you are creating a shift list value equal to its location along the curve. The first section = 0 doesn't get shifted, the second section gets a shift = 1, third = 2, forth = 3 and because the wrap value is set to true the fifth section gets back to 0, sixth = 1 etc etc. creating the twisting effect.
The "one more stupid question" answer is Mass Addition. You will find the component on the Math tab or you can type it into the Keyword search feature (by double clicking the canvas). This component has two outputs a total amount for each list and a partial set of results giving:
List (3,6,9,12)
{0} = 3
{1} = 3+6 = 9
{2} = 3+6+9 = 18
{3} = 3+6+9+12 = 30…