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…
explode curves the data is (0;0;0;0)....(0;2;0;383) and here Arthur Mamou used Path mapper or flip matrix but no one of this solutions is posible in this case because: path mapper extracts (A;B;C;D) and we have items on (A;B;C;D), and flip matrix doesnt work... I have create a subset of list N=4 to get a list with the points of the spheres and flip the matrix to get four branches in order to get the points to input in the pressure force. Sorry but i´m not a teacher and there is, for sure, a better way to do it. Cheers. ; )…
Added by lucas cañada at 4:41pm on January 9, 2013
s:
3
2
1
0
For Simplify I copied this from Grasshopper component help:Simplify a tree by removing the overlap shared amongst all branches. Imagine a tree with six branches: A = {0;1;0}B = {0;1;1}C = {0;1;2}D = {0;2;0}E = {0;2;1}F = {0;3;0}As you can see all these branches share the same first index (0). Thus if you were to Simplify this tree, you'd end up with: A = {1;0}B = {1;1}C = {1;2}D = {2;0}E = {2;1}F = {3;0}…
0} {2;0} {3;0} etc and make them all {0}
S(1) = *;1
R(1) = 1
* is a wildcard therefore Find any branches with 1 as the last branch eg {0;1} {1;1} {2;1} {3;1} etc and make them all {1}
etc etc…