o now is select and group together list items by their index. {0:0} and {2:0} also {1:0} and {3:0}. They way I'm doing this at the moment is quite complicated and I'm sure there's an easier way to achieve the same result.
maybe someone in here knows a solution?
thanks a lot…
English.
Hi all!
I created .ghx for Parametric Design of Bottle. I defined profile curves giving control points with slider, definition of hight, shifting, and offset based on referenced…
What are B and C supposed to be? I mean, they have to resolve to actual numbers.
So given the tree:
{0} [0;1;2;3]
{1} [0;1;2;3]
{2} [0;1;2;3]
{3} [0;1;2;3]
What is it you're after?
Added by David Rutten at 8:59am on October 21, 2017
- False
{0;2}
0 - False
1 - False
2 - True
3 - False
{0;3}
0 - False
1 - False
2 - False
3 - False
Is there anyway I can get a GATE OR result from this list, where the GATE OR is applied between all the corresponding list items. So the end result will be;
{0;0;0} ??
0 - True
1 - True
2 - True
3 - False…
Added by Andrew Butler at 11:26am on November 6, 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…