0;3} (N = 2)
{0;0;0;4} (N = 2)
{0;0;1;0} (N = 2)
{0;0;1;1} (N = 2)
{0;0;1;2} (N = 2)
{0;0;1;3} (N = 2)
{0;0;1;4} (N = 2)
Flattening this structure using the Flatten component would result in:
{0} (N = 20)
However, using a Path Mapper with the following masks will flatten is somewhat more intelligently:
{A;B;C;D} -> {A;B;C}
Now, you get:
{0;0;0} (N = 10)
{0;0;1} (N = 10)
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 3:19am on December 14, 2009
middle index, and choose that point with List Item. If even, for example 4 points (0, 1, 2, 3), you'll get 2, so subtract one and choose those two indices, 1 and 2. I only had a few minutes to play with this, so it isn't a fully-baked solution, but it should take you a little further.…
one list by -1 while disabling index wrapping. Essentially, this removes the first item from one list and the last item from the other list. Conceptually, these are your lists:
Source List: [0, 1, 2, 3, 4, 5]
Shift List 1: [0, 1, 2, 3, 4]
Shift List 2: [1, 2, 3, 4, 5]
If you input these lists into a Line component, you will connect each respective list item with the other, essentially 0-1, 1-2, 2-3, etc.
…
4}
{0;2;0}
{0;2;1}
{0;2;2}
{0;2;3}
{0;2;4}
You cannot flip this because this is more complex than a rectangular matrix. You're going to have to do the mapping yourself. Try a Path Mapper with the following masks:
{A;B;C}(i) -> {A;B;i}(C)
Which should give you a structure that results in 3 lofts.
--
David Rutten
david@mcneel.com
Poprad, Slovakia
…
Added by David Rutten at 3:18pm on November 27, 2011