ents will do or which components will be available.
My problem arises because I want to obtain a list such as the following:
{{6, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6}, {5, 6, 5, 4, 3, 2, 1, 2, 3, 4, 5}, {4, 5, 6, 5, 4, 3, 2, 1, 2, 3, 4}, {3, 4, 5, 6, 5, 4, 3, 2, 1, 2, 3}, {2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 2}, {1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1}}
Which displayed as a matrix is:
If it were possible to combine GH operations (series, shift list, replace string...) with matrices I think it would be quite powerful. A matrix to list component like those available on scientific calculators, would then translate the matrix to list.
For me, matrices come in handy when dealing with surface patterns.
…
Added by Jesus Galvez at 6:46am on November 26, 2012
identical knots on end ends of the curve, so the knot vector will look something like:
0,0,0,1,2,2,2
So one at each end and one in the middle.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 3:47am on January 7, 2013
ee 3)
{5}
0 15
{6}
0 16
And I want to place points at every possible combination of these coordinates, treating Tree 1 as X coordinates, Tree 2 as Y coordinates, and Tree 3 as Z coordinates. Also, I would like the list of points to be a tree with paths corresponding to the coordinates. Wouldn't it be nice if I could plug these trees into a Point XYZ, with a new "branch cross reference" method, and get the following result?
{0:3:5}
0 {10.0, 13.0, 15.0}
{0:3:6}
0 {10.0, 13.0, 16.0}
{0:4:5}
0 {10.0, 14.0, 15.0}
{0:4:6}
0 {10.0, 14.0, 16.0}
{1:3:5}
0 {11.0, 13.0, 15.0}
{1:3:6}
0 {11.0, 13.0, 16.0}
{1:4:5}
0 {11.0, 14.0, 15.0}
{1:4:6}
0 {11.0, 14.0, 16.0}
{2:3:5}
0 {12.0, 13.0, 15.0}
{2:3:6}
0 {12.0, 13.0, 16.0}
{2:4:5}
0 {12.0, 14.0, 15.0}
{2:4:6}
0 {12.0, 14.0, 16.0}
In this form of cross referencing, every combination of individual branches from the different lists is used as separate input, and the output for each combination is put onto a branch in the result whose path is the concatenation of the input branch paths used.…
Added by Andy Edwards at 7:03pm on November 3, 2009
output will show a tree with 3 branches of 4 integers each that I can pass on to other components. What is the best way to do it?
I have tried creating a tree and using a for loop to do so, but it didn't work.
Thank you for your help.
…
shift. I realize I can use 'replace branch' but I do not have an available mask to utilize. I have simplified the problem to its simplest form so my question is understandable, however, the tree I am trying perform this operation on is a much larger 3 digit path address.
{1;3;2}
{2;3;4}
{3;5;4}
{4;3;7}
Change the above list to the list below.
{0;3;2}
{1;3;4}
{2;5:4}
{3;3;7}
I wish for a more robust arsenal of branch manipulation components. Most of the things I need to do are possible with the existing components, however, many operations take several components to perform even simple manipulations. Since branch/path manipulation is so integral to using GH successfully, it seems the GH community would be well served by enhancing the available path manipulation components.
Thanks,
Stan
…