5922
1 - 49888
2 - 70494
3 - 89584
Essentially {0=[0], 1=[0+1], 2=[0+1+2], 3=[0+1+2+3] etc.}
Is there a smart way to do this using GH components?
Thanks!
…
Change the last line to:
Dim CutPlane As New OnNurbsSurface(RhUtil.RhinoCreateSurfaceFromCorners(CornerPts(0), CornerPts(1), CornerPts(2), CornerPts(3)))
For example.
If you have two lists of points.
List A List B
{0;0;0}(0) {0;0}(0)
{0;0;1}(0) {0;1}(0)
{0;2}(0)
{0;3}(0)
{0;4}(0)
And you want to merge the two lists so that the two points in list A are the end points.
Merge Lists Results:
{0;0}(0)
{0;0;0}(0)
{0;0;1}(0)
{0;1}(0)
{0;2}(0)
{0;3}(0)
{0;4}(0)
Because of their path structures the order is wrong from a simple merge so Flattening now is out of the question.
Path Mapper
{A;B} --> {A;B+1}
{A;B;C} --> {A;C*6}
---------------------
Results:
{0;0} --> {0;0+1} = {0;1}
{0;1} --> {0;1+1} = {0;2}
{0;2} --> {0;2+1} = {0;3}
{0;3} --> {0;3+1} = {0;4}
{0;4} --> {0;4+1} = {0;5}
{0;0;0} --> {0;0*6} = {0;0}
{0;0;1} --> {0;1*6} = {0;6}
Now with the Path Structures similar when they are re-ordered the results will have the two points of list A as the end points.
Question 2
why did the curve-line intersection lose the path structure? Both trees had 38 branches.
Both trees had 38 Paths but Tree A had more Items, 147 compared to 38 in Tree B.
So you get this happening:
{0;0;0;0;0;0}(0) compared to {0;0;0;0}(0) results: Null {0;0;0;0;0;0}(0)
Base result paths on longest
{0;0;1;0;0;0}(0) compared to {0;0;0;1}(0) results: Null {0;0;1;0;0;0}(0)
{0;0;2;0;0;0}(0) compared to {0;0;0;2}(0) results: Yes {0;0;2;0;0;0;0}(0)
Add a branch to contain result
{0;0;3;0;0;0}(0) compared to {0;0;0;3}(0) results: Yes {0;0;3;0;0;0;0}(0)
{0;0;3;0;0;0}(1) compared to {0;0;0;3}(0) results: No {0;0;3;0;0;0;1}(0)
{0;0;4;0;0;0}(0) compared to {0;0;0;4}(0) results: Yes {0;0;4;0;0;0;0}(0)
{0;0;4;0;0;0}(1) compared to {0;0;0;4}(0) results: Yes {0;0;4;0;0;0;1}(0)
{0;0;5;0;0;0}(0) compared to {0;0;0;5}(0) results: Yes {0;0;5;0;0;0;0}(0)
{0;0;5;0;0;0}(1) compared to {0;0;0;5}(0) results: Yes {0;0;5;0;0;0;1}(0)
{0;0;5;0;0;0}(2) compared to {0;0;0;5}(0) results: Yes {0;0;5;0;0;0;2}(0)
...... etc
…
(without flatten on inputs): {0;0;0;0}, {0;1;0;3}, {0;2;1;0}
WISH 2 - when using the "only draw geometry for selected objects" mode - then the objects in the custom preview component would show with their custom setting instead of the usual selected objects color (usually green).
…
would like to group the paths based on their item count (n) values resulting in a tree which should look something like this:
{0;0} (3)
{0;1} (2)
{0;2} (2)
{0;3} (1)
in other words, all paths with 2 items are under one path, all with 6 items in another, and so on.
I feel that the pathmapper should be able to do this very easily but cannot figure out what the expression should be... I have tried searching the forum but have not had much luck!
Any ideas? Thanks a ton!…
3} N=2
....
{527} N=4
What I want to do is figure out the same the item in group A and retrieve their branch index so that I can get the corresponding item in group A. Is there any way to make it?…