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. "
except in data, the branches start at 0, so 'group 1' is branch 0
as for the order of your points, that depends on the input prior sorting...
yrs …
0, 5, 10, 15, 20
1, 6, 11, 16, 21
2, 7, 12, 17, 22
3, 8, 13, 18, 23
4, 9, 14, 19, 24
and if i'm here is because i'm not able... :)
can you help me?
thank you
…
13 5 15 6 17 7 ... …
But it seems I have not been clear:
the input values (3,5,7,9,11,13) should be presented in the way of "number slider" parameter. So by increasing the value of the slider from minimum 3, to 5, 7, 9, 13, 15, 17... I should get an output numbers of 0,1,3,4,5,6,7...
So I if choose value 3 on the input "number slider", on the other end, I want 0 as an output.
Or if I choose value 5 on the input "number slider", I want 1 as an output, and so on.
Is this possible?…
ems in the same way. Lofting was particularly difficult, you had to have a separate loft component for every lofted surface that you wanted to generate because the component would/could only see one large list of inputs. Then came along the data structures in GH v0.6 which allowed for the segregation of multiple input sets.
If you go to Section 8: The Garden of Forking Paths of the Grasshopper Primer 2nd Edition you will find the image above describing the storing of data.
Here you will notice a similarity between the path {0;0;0;0}(N=6) and the pathmapper Mask {A;B;C;D}(i). A is a placeholder for all of the first Branch structures (in this case just 0). B is a place holder for all the second branch structures possibly either 0, 1 or 2 in this case. And so forth.
(i) is a place holder for the index of N. If you think of it like a for loop the i plays the same role. For the example {A;B;C;D}(i) --> {i\3}
{0;0;0;0}(0) --> {0\3} = {0}
{0;0;0;0}(1) --> {1\3} = {0}
{0;0;0;0}(2) --> {2\3} = {0}
{0;0;0;0}(3) --> {3\3} = {1}
{0;0;0;0}(4) --> {4\3} = {1}
{0;0;0;0}(5) --> {5\3} = {1}
{0;0;0;1}(0) --> {0\3} = {0}
{0;0;0;1}(1) --> {1\3} = {0}
{0;0;0;1}(2) --> {2\3} = {0}
{0;0;0;1}(3) --> {3\3} = {1}
{0;0;0;1}(4) --> {4\3} = {1}
{0;0;0;1}(5) --> {5\3} = {1}
{0;0;0;1}(6) --> {6\3} = {2}
{0;0;0;1}(7) --> {7\3} = {2}
{0;0;0;1}(8) --> {8\3} = {2}
...
{0;2;1;1}(8) --> {8\3} = {2}
I'm not entirely sure why you want to do this particular exercise but it goes some way towards describing the process.
The reason for the tidy up: every time the data stream passes through a component that influences the path structure it adds a branch. This can get very unwieldy if you let it go to far. some times I've ended up with structures like {0;0;1;0;0;0;3;0;0;0;14}(N=1) and by remapping the structure to {A;B;C} you get {0;0;1}(N=15) and is much neater to deal with.
If you ever need to see what the structure is there is a component called Param Viewer on the first Tab Param>Special Icon is a tree. It has two modes text and visual double click to switch between the two.
Have a look at this example of three scenarios in three situations to see how the data structure changes depending on what components are doing.
…
o relate to each plane its sections i.e.:
paths=624 - {0;0;2;0}(N=1)
the number of section planes is 70 and somewhere I have 4 section lines and somewhere 5 or 8, is it clear?
1 plane = 1 set of sections
thank you for helping me.…
Added by paul piciorul at 5:35am on October 28, 2010
t 2^15 are being listed. Later I will use this info to cull;
all instances where important nodes are not present,
geometric unstable trusses using Grubler's criterion,
the remainder will go to FEM for further analysis,
I can use this method but i require to separate each binary boolean
so they are stored separately like so;
{0} (0) 0 (1) 0 (2) 0 (3) 0 (4) 0 . . . . . . . . (15) 0
{1} (0) 1 (1) 0 (2) 0 (3) 0 (4) 0 . . . . . . . . (15) 0
{2} (0) 0 (1) 1 (2) 0 (3) 0 (4) 0 . . . . . . . . (15) 0 (I know there must be an easier scripted way how to do it) or else is there a way to decompose them directly from the method you sent me using standard Grasshopper components?. Any method would work.
Best,
Kane…